mm-4709 === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? Am 30.06.2008 13:48 schrieb Gottfried Helms: > Am 30.06.2008 12:08 schrieb alainverghote@gmail.com: Well, from the notepad: (e^x -1)/2 = tfÁ0(x) - tfÁ1(x) + tfÁ2(x) - ... +... (e^tf(x) -1)/2 = tfÁ1(x) - tfÁ2(x) + tfÁ3(x) - ... +... (exp(tf(x)) -1)/2 + (exp(x) -1)/2 = x exp(tf(x)) -1 + exp(x) -1 = 2x exp(tf(x)) = 2x + 2 - exp(x) tf(x) = log( 2x + 2 - exp(x)) Now, what does g(x) = asn(x) look like? g(x) = tfÁ0(x) - tfÁ-1(x) + tfÁ-2(x) - ... +... g(tf(x)) = tfÁ-1(x) - tfÁ-2(x) + tfÁ3(x) - ... +... g(tf(x)) + g(x) = x g(log(2 + 2x - exp(x))) + g(x) = x How to proceed? Hmmm... Gottfried === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? > Am 30.06.2008 13:48 schrieb Gottfried Helms: > Am 30.06.2008 12:08 schrieb alainvergh...@gmail.com: Well, from the notepad: æ(e^x -1)/2 æ æ = tfÁ0(x) - tfÁ1(x) + tfÁ2(x) - ... +... > æ(e^tf(x) -1)/2 = æ æ æ æ æ tfÁ1(x) - tfÁ2(x) + tfÁ3(x) - ... +... æ(exp(tf(x)) -1)/2 + (exp(x) -1)/2 = x æexp(tf(x)) -1 + exp(x) -1 = 2x > æexp(tf(x)) = 2x + 2 - exp(x) > ætf(x) = log( 2x + 2 - exp(x)) Now, what does g(x) = asn(x) look like? æg(x) æ æ = tfÁ0(x) - tfÁ-1(x) + tfÁ-2(x) - ... +... > æg(tf(x)) = æ æ æ æ æ tfÁ-1(x) - tfÁ-2(x) + tfÁ3(x) - ... +... æg(tf(x)) + g(x) = x > æg(log(2 + 2x - exp(x))) + g(x) = x æHow to proceed? > æHmmm... Gottfried Well, inverse function of f(x) using lambertW() is -Lambert(-1/2exp(1/2e^x - 1)) +1/2e^x - 1 series near 0 , x +x^2 +4/3x^3 +29/12x^4 +51/10x^5 ... Alain === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? Am 30.06.2008 17:00 schrieb alainverghote@gmail.com: > Gottfried Well, inverse function of f(x) using lambertW() > is -Lambert(-1/2exp(1/2e^x - 1)) +1/2e^x - 1 > series near 0 , x +x^2 +4/3x^3 +29/12x^4 +51/10x^5 ... Yepp, I got the same series - good! But now - iterations and especially the sum of iterations of this functions in the Lambert-representation should be intractable, so I don't assume this can be helpful to get more insight in the source of the inconsisteny-problem; remember: by the formal application of the matrix-approach asn(x)+asp(x)-x = 0 // expected or asn(x,f(x)) = x - asp(x,f(x)) // expected or asp(x,fÁ(-1)(x)) = x - asp(x,f(x)) // expected is not true, at least for the function f(x) = exp(x)-1 Well - it was a try... -------------------------------------- Meanwhile I refined my computation-process, so I've now even the function fz(x) with the condition e^x - 1 = fz(x) + fz(fz(x)) + fz(fz(fz(x))) + .. = sum{h=1..inf} fzÁh(x) I got fz(x) = 2*(x/4)/1! + 6*(x/4)^2/2! + 10*(x/4)^3/3! - 46*(x/4)^4/4! - 554*(x/4)^5/5! - 1690*(x/4)^6/6! + 27882*(x/4)^7/7! + 505986*(x/4)^8/8! + 2529590*(x/4)^9/9! - 61918794*(x/4)^10/10! - 1726391798*(x/4)^11/11! - 14268435022*(x/4)^12/12! + 352044609814*(x/4)^13/13! + O(x^14) where the integer parts of the coefficients are [2] [6] [10] [-46] [-554] [-1690] [27882] [505986] [2529590] [-61918794] [-1726391798] [-14268435022] [352044609814] which have to be divided py powers of 4 and by factorials to give the coefficients of the function. The float representation of this function is fz(x) = 0.500000000000*x^1 + 0.187500000000*x^2 + 0.0260416666667*x^3 - 0.00748697916667*x^4 - 0.00450846354167*x^5 - 0.000573052300347*x^6 + 0.000337655203683*x^7 + 0.000191486449469*x^8 + 0.0000265917660278*x^9 - 0.0000162726971838*x^10 - 0.0000103115449999*x^11 - 0.00000177549511523*x^12 + 0.000000842437121499*x^13 + 0.000000632647393830*x^14 + O(x^15) Can we give a range for x where this converges ? The quotients of subsequent coefficients give the following sequence 0.375000000000,0.138888888889,-0.287500000000, 0.602173913043,0.127105896510,-0.589222316145, 0.567106466538,0.138870223462,-0.611944959460, 0.633671534805,0.172185168687,-0.474480112208, 0.750972835462,0.212965249501,-0.333234112499, 0.928624638217,0.256835516535,-0.218432174828, 1.22327362079,0.303250445581,-0.126531683586, 1.82969444504,0.353026162019,-0.0509774937397, 3.94499797418,0.407780103462,0.0133041813614,-13.1183083394, 0.470026310618,0.0698970628404,-2.15481130004, 0.543690120454,... ----------------------------------------------------------- Using 32 coefficients for the function and 60 iterates for the sum I could approximate e^1 -1 relatively well. I got sum(h=1,60,fzÁh(1.0)) - ( exp(1)-1 ) = -3.24385306514 E-13 where the quality of approximation increased when terms of the function and numbers of iterates are increased. Fun... :-) Gottfried Helms === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? <6crvh8F3ikr03U1@mid.dfncis.de>6cs0dqF3gobboU1@mid.dfncis.de> <6d11bqF9nqlU1@mid.dfncis.de> posting-account=06BQLAoAAADoC7Y4z9FWcUwGvMa7xMG9 7.4),gzip(gfe),gzip(gfe) > Am 30.06.2008 17:00 schrieb alainvergh...@gmail.com:> Gottfried > Well, inverse function of f(x) using lambertW() > is æ-Lambert(-1/2exp(1/2e^x - 1)) +1/2e^x - 1 > series near 0 , x +x^2 +4/3x^3 +29/12x^4 +51/10x^5 ... Yepp, I got the same series - good! But now - iterations and especially the sum of iterations > of this functions in the Lambert-representation should be > intractable, so I don't assume this can be helpful to get > more insight in the source of the inconsisteny-problem; remember: > by the formal application of the matrix-approach > æ æ æ asn(x)+asp(x)-x = 0 æ// expected > or > æ æ æ asn(x,f(x)) = x - asp(x,f(x)) // expected > or > æ æ æ asp(x,fÁ(-1)(x)) = x - asp(x,f(x)) // expected is not true, at least for the function f(x) = exp(x)-1 Well - it was a try... -------------------------------------- Meanwhile I refined my computation-process, so I've > now even the function fz(x) with the condition æ e^x - 1 = fz(x) + fz(fz(x)) + fz(fz(fz(x))) + .. > æ æ æ æ æ = sum{h=1..inf} fzÁh(x) I got > æfz(x) = 2*(x/4)/1! + 6*(x/4)^2/2! + 10*(x/4)^3/3! - 46*(x/4)^4/4! - 554*(x/4)^5/5! > æ æ æ æ- 1690*(x/4)^6/6! + 27882*(x/4)^7/7! + 505986*(x/4)^8/8! + 2529590*(x/4)^9/9! > æ æ æ æ- 61918794*(x/4)^10/10! - 1726391798*(x/4)^11/11! - 14268435022*(x/4)^12/12! > æ æ æ æ+ 352044609814*(x/4)^13/13! + O(x^14) where the integer parts of the coefficients are > [2] > [6] > [10] > [-46] > [-554] > [-1690] > [27882] > [505986] > [2529590] > [-61918794] > [-1726391798] > [-14268435022] > [352044609814] > which have to be divided py powers of 4 and by factorials to give the > coefficients of the function. The float representation of this function is > æfz(x) = æ 0.500000000000*x^1 + 0.187500000000*x^2 + 0.0260416666667*x^3 - 0.00748697916667*x^4 > æ æ æ æ æ- 0.00450846354167*x^5 - 0.000573052300347*x^6 + 0.000337655203683*x^7 + 0.000191486449469*x^8 > æ æ æ æ æ+ 0.0000265917660278*x^9 - 0.0000162726971838*x^10 - 0.0000103115449999*x^11 > æ æ æ æ æ- 0.00000177549511523*x^12 + 0.000000842437121499*x^13 + 0.000000632647393830*x^14 > æ æ æ æ æ+ O(x^15) Can we give a range for x where this converges ? > The quotients of subsequent coefficients give the following sequence æ0.375000000000,0.138888888889,-0.287500000000, > æ0.602173913043,0.127105896510,-0.589222316145, > æ0.567106466538,0.138870223462,-0.611944959460, > æ0.633671534805,0.172185168687,-0.474480112208, > æ0.750972835462,0.212965249501,-0.333234112499, > æ0.928624638217,0.256835516535,-0.218432174828, > æ1.22327362079,0.303250445581,-0.126531683586, > æ1.82969444504,0.353026162019,-0.0509774937397, > æ3.94499797418,0.407780103462,0.0133041813614,-13.1183083394, > æ0.470026310618,0.0698970628404,-2.15481130004, > æ0.543690120454,... ----------------------------------------------------------- Using 32 coefficients for the function and 60 iterates for the sum > I could approximate e^1 -1 relatively well. I got æ æ sum(h=1,60,fzÁh(1.0)) - ( exp(1)-1 ) = æ-3.24385306514 E-13 where the quality of approximation increased when terms of the > function and numbers of iterates are increased. Fun... :-) Gottfried Helms But the last function you've arrived at is : -LambertW(exp(exp(x)) + exp(x) series near 0 , 1/2x +3/16x^2 +5/192x^3 - ... alain === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? Am 02.07.2008 14:43 schrieb alainverghote@gmail.com: But the last function you've arrived at is : > -LambertW(exp(exp(x)) + exp(x) > series near 0 , 1/2x +3/16x^2 +5/192x^3 - ... > Well - that's beginning to get true fun... :-) Let's go back to the previous function and its inverse. f(x)=ln( -exp(x) +2*x+2) > Well, inverse function of f(x) using lambertW() > is -Lambert(-1/2exp(1/2e^x - 1)) +1/2e^x - 1 so g(x) = -Lambert(-1/2exp(1/2e^x - 1)) +1/2e^x - 1 and g(x) = fÁ(-1)(x) Now, for the alternating sum of iterates of f(x) asp(x) = x - f(x) + fÁ2(x) - fÁ3(x) + ... - ... and of g(x) asn(x) = x - g(x) + gÁ2(x) - gÁ3(x) + ... - ... where for asp(x) we have the closed form asp(x) = (exp(x)-1)/2 it would be good to have a closed form for asn(x) as well. This seems out of reach. Could Mathematica/Maple or such be made to compute asn(x) by evaluation of the iterates and summing these terms? I assume, it is in the near of x - (e^x-1)/2 but not exactly. If it is exactly the same, then some aspect of divergence must be interfering the summation of the lambertW-representation of g(x) and its iterates. (Unfortunately I don't have access to Mathematica/ Maple and its lambertw-implementation) Gottfried === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? Am 02.07.2008 20:00 schrieb Gottfried Helms: > Now, for the alternating sum of iterates of f(x) asp(x) = x - f(x) + fÁ2(x) - fÁ3(x) + ... - ... and of g(x) asn(x) = x - g(x) + gÁ2(x) - gÁ3(x) + ... - ... where for asp(x) we have the closed form > asp(x) = (exp(x)-1)/2 > it would be good to have a closed form for asn(x) > as well. This seems out of reach. > I just tried with a poor man's lambertW-implementation. To have convergence/summability I used the parameter x=-0.05 using the closed-form for alternating sum of f-iterates: asp(-0.05) = (exp(-0.05)-1)/2 = -0.0243852877496 using alternating sum of g-iterates: asn(-0.05) = sumalt(k=0,(-1)^k*g(-0.05,k)) = -0.0243852877496 The difference: asn(-0.05) - asp(-0.05) = 2.11329439961 E-99 ... which is again surprising, and the test of the conjecture of zero-value asn(-0.05) + asp(-0.05) -(-0.05) = 0.00122942450071 =/= 0 is negative, so indeed the serial summation of terms seems seem to give the correct result - fine. That the difference of asn and asp is near zero remains now as another surprise... Gottfried Helms === Subject: Re: -- Tetration: x - f(x) + f(f(x)) - f(f(f(x))) + ... - ... = 1/2*(exp(x) - 1) ?? / Error! Am 03.07.2008 07:34 schrieb Gottfried Helms: > I just tried with a poor man's lambertW-implementation. > To have convergence/summability I used the parameter x=-0.05 using the closed-form for alternating sum of f-iterates: asp(-0.05) = (exp(-0.05)-1)/2 = -0.0243852877496 I made an error in the computation of the sum, giving crap. I'll repost corrected results soon. Gottfried Helms === Subject: Re: MinMax Technology > You're welcome. You've hit upon one of my favorite interests, namely > order theory. It's exceeding more complex than what you'd expect from > the simplicity of (partial) orders. I've been revising my notes upon > ordering a partition of an ordered set and it's applications to > lattices, ordered groups and lattice ordered groups. Maybe I'll get it > together enough to present to the group. Again, I'll be wanting to know > why it too is not a counter example to William's Metatheorm: > Whatever math I dream up is already old hat. Let's summarize again the laws of MinMax Technology. William Elliot is quite right when he says that the system presented is not really a NFL (Negative Fuzzy Logic), in the sense that the complements (-x) are not in the same logic; as I've defined it. But anyway, below are the rules. Associativity: min(a,min(b,c) = min(min(a,b),c) max(a,max(b,c) = max(max(a,b),c) Commutativity: min(a,b) = min(b,a) max(a,b) = max(b,a) Distributivity: min(a,max(b,c) = max(min(a,b),min(a,c)) max(a,min(b,c) = min(max(a,b),max(a,c)) Absorption: max(a,min(a,b) = a min(a,max(a,b) = a Idempotence: min(a,a) = a max(a,a) = a De Morgan's laws: -min(a,b) = max(-a,-b) -max(a,b) = min(-a,-b) a <= b : a = min(a,b) if and only if max(a,b) = b MinMax Technology shows kind of a split between positive and negative as we arrive at the following properties: For x <= 0 : min(x,0) = x , max(x,0) = 0 min(-x,0) = 0 , max(-x,0) = -x min(x,-x) = x , max(x,-x) = -x For x >= 0 : min(x,0) = 0 , max(x,0) = x min(-x,0) = -x , max(-x,0) = 0 min(x,-x) = -x , max(x,-x) = x It cannot be denied that there is some resemblance between the above MinMax algebra - or whatever it should be called - and the rules of common Boolean logic. There is an even closer relationship with the rules of Fuzzy Logic. But these analogies are not quite unambiguous, though I wish they were. The ambiguity is in the following choices: - does an intersection / correspond with a Min or with a Max ? - if then does a union / correspond with a Max or with a Min ? In particular, we have the formula for being approximately a subset: X <= Y iff /_i /_j (x_i = y_j) where i=1,..,M and j=1,..,N Within the original posting, we have translated this formula into the language of real numbers by equivalencing (x_i = y_j) with a distance, equivalencing / with a product, and equivalencing / with a sum of squares. It is noted that equivalencing / with a product is exactly the other way around as is common in Fuzzy Logic. Which is troublesome. The right formula in MinMax Technology is the other way around as well: X <= Y iff max_i min_j ( |x_i - y_j| ) = residual (positive) Which is not in concordance with common Fuzzy Logic, as it would read: X <= Y iff min_i max_j ( |x_i - y_j| ) = residual (positive) But if there does not exist a consistent Negative Fuzzy Logic then any reason behind the good choice becomes questionable. Do we have to make a right and a wrong choice then ? And let practice decide which one to single out ? Like have a positive and a negative root with a quadratic equation and let the positive root survive from certain considerations outside theory .. Not quite satisfactory if a viewpoint doesn't fit in some well established tradition, somehow, somewhere .. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl > Let's summarize again the laws of MinMax Technology. William Elliot is > quite right when he says that the system presented is not really a NFL > (Negative Fuzzy Logic), in the sense that the complements (-x) are not > in the same logic; as I've defined it. But anyway, below are the rules. Associativity: min(a,min(b,c) = min(min(a,b),c) > max(a,max(b,c) = max(max(a,b),c) > Commutativity: min(a,b) = min(b,a) > max(a,b) = max(b,a) > Distributivity: min(a,max(b,c) = max(min(a,b),min(a,c)) > max(a,min(b,c) = min(max(a,b),max(a,c)) > Absorption: max(a,min(a,b) = a > min(a,max(a,b) = a Idempotence: min(a,a) = a > max(a,a) = a > De Morgan's laws: -min(a,b) = max(-a,-b) > -max(a,b) = min(-a,-b) a <= b : a = min(a,b) if and only if max(a,b) = b > Call the pertinent set S. (S,min,max) is a distribuitive lattice. (S,min,max,-) is a lattice with an additional unary operator. > MinMax Technology shows kind of a split between positive and negative > as we arrive at the following properties: For x <= 0 : min(x,0) = x , max(x,0) = 0 > min(-x,0) = 0 , max(-x,0) = -x > min(x,-x) = x , max(x,-x) = -x For x >= 0 : min(x,0) = 0 , max(x,0) = x > min(-x,0) = -x , max(-x,0) = 0 > min(x,-x) = -x , max(x,-x) = x > No you don't. You're adding the asiom -0 = 0. > It cannot be denied that there is some resemblance between the above > MinMax algebra - or whatever it should be called - and the rules of > common Boolean logic. There is an even closer relationship with the > rules of Fuzzy Logic. But these analogies are not quite unambiguous, > though I wish they were. The ambiguity is in the following choices: > It's looking much like a lattice ordered group. > - does an intersection / correspond with a Min or with a Max ? > - if then does a union / correspond with a Max or with a Min ? > (P(S), /, /), ie (P(S), intersection, uninn), is the lattice of subsets S. Clearly A / B subset A,B subset A / B. > In particular, we have the formula for being approximately a subset: X <= Y iff /_i /_j (x_i = y_j) where i=1,..,M and j=1,..,N > A subset B when for all a in A, a in B iff for all a in A, some b in B with a = b iff for all a in A, OR_(b in B) (a = b) iff AND_(a in A) OR_(b in B) (a = b) Ok, got it this time. Have fun with infinite statements. There's actually a logic that considers them. > Within the original posting, we have translated this formula into the > language of real numbers by equivalencing (x_i = y_j) with a distance, > equivalencing / with a product, and equivalencing / with a sum of > squares. It is noted that equivalencing / with a product is exactly > the other way around as is common in Fuzzy Logic. Which is troublesome. > The right formula in MinMax Technology is the other way around as well: X <= Y iff max_i min_j ( |x_i - y_j| ) = residual (positive) > If X subset Y, then min_j |x_i - y_j| = 0. X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, Left to right is nearly immediate as pointed out above. Right to left requires that max and min exist. When they don't, you are not at liberty to use inf, sup in lieu of min, max. X = {0}, Y = 1/N = { 1/n | n in N } for example. > Which is not in concordance with common Fuzzy Logic, as it would read: X <= Y iff min_i max_j ( |x_i - y_j| ) = residual (positive) > Let X = { 0 }, Y = { 1 }. min_i max_j |x_i - y_j| = 1 > But if there does not exist a consistent Negative Fuzzy Logic then any > reason behind the good choice becomes questionable. Do we have to make > a right and a wrong choice then ? And let practice decide which one to > single out ? Like have a positive and a negative root with a quadratic > equation and let the positive root survive from certain considerations > outside theory .. Not quite satisfactory if a viewpoint doesn't fit in > some well established tradition, somehow, somewhere .. Han de Bruijn === Subject: Re: MinMax Technology > X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. Quite right ! For that reason, there is a min-max analogue of a (finite) points cloud as approximately a subset of a line, but there is NO minmax analogue of a straight line as approximately a subset of a points cloud. The latter can only be accomplished for a line _segment_ with endpoints. Now that you've clearly got the gist of it, what would be the analogue, in min-max language, of two sets being approximately _disjoint_? This is maybe an important question for clustering techniques .. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <7149c$4868c499$82a1e228$30961@news1.tudelft.nl > X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. > Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. Quite right ! For that reason, there is a min-max analogue of a (finite) > points cloud as approximately a subset of a line, but there is NO minmax > analogue of a straight line as approximately a subset of a points cloud. > The latter can only be accomplished for a line _segment_ with endpoints. Now that you've clearly got the gist of it, what would be the analogue, > in min-max language, of two sets being approximately _disjoint_? This is > maybe an important question for clustering techniques .. > A,B disjoint iff for all a in A, b in B, a /= b. === Subject: Re: MinMax Technology X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. > Quite right ! For that reason, there is a min-max analogue of a (finite) > points cloud as approximately a subset of a line, but there is NO minmax > analogue of a straight line as approximately a subset of a points cloud. > The latter can only be accomplished for a line _segment_ with endpoints. I have a visualization of this at the site now. You could take a look: http://hdebruijn.soo.dto.tudelft.nl/jaar2008/minmax_0.htm > Now that you've clearly got the gist of it, what would be the analogue, > in min-max language, of two sets being approximately _disjoint_? This is > maybe an important question for clustering techniques .. > A,B disjoint iff for all a in A, b in B, a /= b. Mind the approximate; sci.math.num_analysis crossposting on purpose .. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <7149c$4868c499$82a1e228$30961@news1.tudelft.nl> <20080630044243.I56029@agora.rdrop.com> <1deec$4868d8b4$82a1e228$8620@news1.tudelft.nl > X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. > > Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. > Quite right ! For that reason, there is a min-max analogue of a (finite) > points cloud as approximately a subset of a line, but there is NO minmax > analogue of a straight line as approximately a subset of a points cloud. > The latter can only be accomplished for a line _segment_ with endpoints. I have a visualization of this at the site now. You could take a look: > http://hdebruijn.soo.dto.tudelft.nl/jaar2008/minmax_0.htm I lack graphic capacity. > Now that you've clearly got the gist of it, what would be the analogue, > in min-max language, of two sets being approximately _disjoint_? This is > maybe an important question for clustering techniques .. > A,B disjoint iff for all a in A, b in B, a /= b. Mind the approximate; sci.math.num_analysis crossposting on purpose .. Han de Bruijn === Subject: Re: MinMax Technology > I lack graphic capacity. That's a pity. Especially when somebody would like to exchange ideas of _geometry_ with you. Identify X with a line segment, identify Y with a (random) points cloud, both in the Euclidian plane. Consider: (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual Then explain why points in the cloud that fulfill the above condition (Conjecture / Theorem / Excercise :) almost invariably come in _pairs_. http://hdebruijn.soo.dto.tudelft.nl/jaar2008/minmax_0.htm The inverse problem is much less uncommon. Identify X with a points cloud, identify Y with a line segment. Consider again: (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual Then the minimum distance between any point in the cloud and the line segment is given by the length of a perpendicular. Next determine the perpendicular with the greatest length (-> residual) and you're done. Still apart from special cases at the end-points of the segment. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <7149c$4868c499$82a1e228$30961@news1.tudelft.nl> <20080630044243.I56029@agora.rdrop.com> <1deec$4868d8b4$82a1e228$8620@news1.tudelft.nl> False. Let X = Y = {x}. No residual, just zero. > Then explain why points in the cloud that fulfill the above condition > (Conjecture / Theorem / Excercise :) almost invariably come in _pairs_. > Huh? Are you asking us to consider the points for which max_(x in X) min_(y in Y) |x - y| <= epsilon for some small epsilon? Doesn't make sense as the expression has no free x nor free y. Maybe you want to consider the points x in X with min_(y in Y) |x - y| <= epsilon which will include all the points of X that are in Y. > The inverse problem is much less uncommon. Identify X with a points > cloud, identify Y with a line segment. Consider again: (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual > Upon reconsideration I still get the same 'false' answer. Now as the cloud is two or three dimensional, how are we to understand |x - y|? As the absolute difference of two real numbers or the distance between two points x,y in 1,2 or 3 space. > Then the minimum distance between any point in the cloud and the line > segment is given by the length of a perpendicular. Next determine the > perpendicular with the greatest length (-> residual) and you're done. > Still apart from special cases at the end-points of the segment. > The locus of points a given distance from a line seqment is a rectangle with semi-circle ends. To follow your discussion is to understand fuzzy thinking. ---- === Subject: Re: MinMax Technology > Consider: > (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual > False. Let X = Y = {x}. No residual, just zero. Not false but true. A zero residual is just the crisp special case, corresponding with classical set theory; read _approximately_ subset instead of just subset (I thought that was clear from the context). > Then explain why points in the cloud that fulfill the above condition > (Conjecture / Theorem / Excercise :) almost invariably come in _pairs_. > Huh? Are you asking us to consider the points for which > max_(x in X) min_(y in Y) |x - y| <= epsilon > for some small epsilon? Doesn't make sense as the expression > has no free x nor free y. Maybe you want to consider the points > x in X with > min_(y in Y) |x - y| <= epsilon which will include all the points of X that are in Y. No. Calculate max_(x in X) min_(y in Y) |x - y| and find all x and y for which this number is equal to |x - y| . Sorry. Better in this way ? > The inverse problem is much less uncommon. Identify X with a points > cloud, identify Y with a line segment. Consider again: > (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual > Upon reconsideration I still get the same 'false' answer. Same. Calculate max_(x in X) min_(y in Y) |x - y| and find all x and y for which this number is equal to |x - y| . > Now as the cloud is two or three dimensional, how are we to understand > |x - y|? As the absolute difference of two real numbers or the distance > between two points x,y in 1,2 or 3 space. The latter. > Then the minimum distance between any point in the cloud and the line > segment is given by the length of a perpendicular. Next determine the > perpendicular with the greatest length (-> residual) and you're done. > Still apart from special cases at the end-points of the segment. > The locus of points a given distance from a line > seqment is a rectangle with semi-circle ends. Correct. > To follow your discussion is to understand fuzzy thinking. Alas. My humble programming efforts usually result in more clarity than the accompanying theoretical framework. Apologies for this. Some of us don't _know_, but they _do_ it. You comments have been helpful ayway. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <7149c$4868c499$82a1e228$30961@news1.tudelft.nl> <20080630044243.I56029@agora.rdrop.com> <1deec$4868d8b4$82a1e228$8620@news1.tudelft.nl> <20080701015754.O19461@agora.rdrop.com> <7e76e$486a0078$82a1e228$9664@news2.tudelft.nl > Consider: > (X subset Y) iff max_(x in X) min_(y in Y) |x - y| = residual > False. Let X = Y = {x}. No residual, just zero. Not false but true. A zero residual is just the crisp special case, > corresponding with classical set theory; read _approximately_ subset > instead of just subset (I thought that was clear from the context). > No it wasn't. I'm not a mind reader. > Then explain why points in the cloud that fulfill the above condition > (Conjecture / Theorem / Excercise :) almost invariably come in _pairs_. > Huh? Are you asking us to consider the points for which > max_(x in X) min_(y in Y) |x - y| <= epsilon > for some small epsilon? Doesn't make sense as the expression > has no free x nor free y. Maybe you want to consider the points > x in X with > min_(y in Y) |x - y| <= epsilon > > which will include all the points of X that are in Y. No. Calculate max_(x in X) min_(y in Y) |x - y| and find all x and y > for which this number is equal to |x - y| . Sorry. Better in this way ? > A crisp calculation or a fuzzy calculation? Anyway yuu want to find { (x,y) in XxY : |x - y| = max_(x in X) min_(y in Y) |x - y| } X = { 1,2,3 } Y = { 4,5,6 } m4 = 1, m5 = 2, m6 = 3; max = 3 (1,4), (2,5) & (3,6) are a pair? > Now as the cloud is two or three dimensional, how are we to understand > |x - y|? As the absolute difference of two real numbers or the distance > between two points x,y in 1,2 or 3 space. The latter. > ---- === Subject: Re: MinMax Technology X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. > Quite right ! For that reason, there is a min-max analogue of a (finite) > points cloud as approximately a subset of a line, but there is NO minmax > analogue of a straight line as approximately a subset of a points cloud. > The latter can only be accomplished for a line _segment_ with endpoints. > Now that you've clearly got the gist of it, what would be the analogue, > in min-max language, of two sets being approximately _disjoint_? This is > maybe an important question for clustering techniques .. > A,B disjoint iff for all a in A, b in B, a /= b. Yes. But now for approximately the best. Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <7149c$4868c499$82a1e228$30961@news1.tudelft.nl> <20080630044243.I56029@agora.rdrop.com> <20989$4868d700$82a1e228$8058@news1.tudelft.nl > X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. > > Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. > Now that you've clearly got the gist of it, what would be the analogue, > in min-max language, of two sets being approximately _disjoint_? This is > maybe an important question for clustering techniques .. > A,B disjoint iff for all a in A, b in B, a /= b. Yes. But now for approximately the best. > Huh? What does that mean? Cast it into your min max form? A,B disjoint iff AND_(a in A) AND_(b in B) (a /= b) or the slower computation min_(a in A) min_(b in B) |a - b| /= 0 === Subject: Re: MinMax Technology > Huh? What does that mean? Cast it into your min max form? A,B disjoint iff AND_(a in A) AND_(b in B) (a /= b) or the slower computation > min_(a in A) min_(b in B) |a - b| /= 0 That's correct ! Nothing Huh, huh ? Han de Bruijn === Subject: Re: MinMax Technology > A,B disjoint iff for all a in A, b in B, a /= b. Yes. But now for approximately the best. Sigh ! Why do I always have to do everything myself ? Using William Elliot's notation, as in > X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0 the MinMax analogue is: min_(x in X) min_(y in Y) |x - y| <> 0 And the larger the residual, the better disjoint. Applications of the same idea appeared on the internet ('sci.math.*') as the independent threads closest point & best root finding strategy : Han de Bruijn === Subject: Re: MinMax Technology > In particular, we have the formula for being approximately a subset: > X <= Y iff /_i /_j (x_i = y_j) where i=1,..,M and j=1,..,N > A subset B when for all a in A, a in B > iff for all a in A, some b in B with a = b > iff for all a in A, OR_(b in B) (a = b) > iff AND_(a in A) OR_(b in B) (a = b) Ok, got it this time. Have fun with infinite statements. > There's actually a logic that considers them. I'm eager to learn. Can you provide references? Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <2e477$4868c0e7$82a1e228$16436@news1.tudelft.nl > In particular, we have the formula for being approximately a subset: X <= Y iff /_i /_j (x_i = y_j) where i=1,..,M and j=1,..,N > A subset B when for all a in A, a in B > iff for all a in A, some b in B with a = b > iff for all a in A, OR_(b in B) (a = b) > iff AND_(a in A) OR_(b in B) (a = b) > Ok, got it this time. Have fun with infinite statements. > There's actually a logic that considers them. I'm eager to learn. Can you provide references? Han de Bruijn > No. Perhaps Yahoo infinite logic or Wikipedia about logic. === Subject: Re: MinMax Technology > MinMax Technology shows kind of a split between positive and negative > as we arrive at the following properties: > For x <= 0 : min(x,0) = x , max(x,0) = 0 > min(-x,0) = 0 , max(-x,0) = -x > min(x,-x) = x , max(x,-x) = -x > For x >= 0 : min(x,0) = 0 , max(x,0) = x > min(-x,0) = -x , max(-x,0) = 0 > min(x,-x) = -x , max(x,-x) = x > No you don't. You're adding the asiom -0 = 0. Come on, William! There is only common real numbers arithmetic here! So the 0 simply denotes a zero and -x simply denotes minus x . No asioms (you mean: axioms) added anywhere. The system behaves differently for + and - numbers, right? Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> <4a794$4868c060$82a1e228$16436@news1.tudelft.nl > MinMax Technology shows kind of a split between positive and negative > as we arrive at the following properties: For x <= 0 : min(x,0) = x , max(x,0) = 0 > min(-x,0) = 0 , max(-x,0) = -x > min(x,-x) = x , max(x,-x) = -x For x >= 0 : min(x,0) = 0 , max(x,0) = x > min(-x,0) = -x , max(-x,0) = 0 > min(x,-x) = -x , max(x,-x) = x > No you don't. You're adding the asiom -0 = 0. Come on, William! There is only common real numbers arithmetic here! So > the 0 simply denotes a zero and -x simply denotes minus x . No asioms > (you mean: axioms) added anywhere. The system behaves differently for + > and - numbers, right? > In your revision of the rules you didn't say anything about the underlying set. Besides, if you're wanting to generalize, then seemly you wouldn't want to limit your ideas to the one-dimensional real line === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <39e2d$486893b7$82a1e228$29310@news1.tudelft.nl> <20080630022127.R31231@agora.rdrop.com> Let's summarize again the laws of MinMax Technology. William Elliot is > quite right when he says that the system presented is not really a NFL > (Negative Fuzzy Logic), in the sense that the complements (-x) are not > in the same logic; as I've defined it. But anyway, below are the rules. > Associativity: min(a,min(b,c) = min(min(a,b),c) > max(a,max(b,c) = max(max(a,b),c) > Commutativity: min(a,b) = min(b,a) > max(a,b) = max(b,a) > Distributivity: min(a,max(b,c) = max(min(a,b),min(a,c)) > max(a,min(b,c) = min(max(a,b),max(a,c)) > Absorption: max(a,min(a,b) = a > min(a,max(a,b) = a > Idempotence: min(a,a) = a > max(a,a) = a > De Morgan's laws: -min(a,b) = max(-a,-b) > -max(a,b) = min(-a,-b) > a <= b : a = min(a,b) if and only if max(a,b) = b > Exercise (using the above rules only): a <= b iff -b <= -a > Call the pertinent set S. (S,min,max) is a distribuitive lattice. > (S,min,max,-) is a lattice with an additional unary operator. > MinMax Technology shows kind of a split between positive and negative > as we arrive at the following properties: > For x <= 0 : min(x,0) = x , max(x,0) = 0 > min(-x,0) = 0 , max(-x,0) = -x > min(x,-x) = x , max(x,-x) = -x > For x >= 0 : min(x,0) = 0 , max(x,0) = x > min(-x,0) = -x , max(-x,0) = 0 > min(x,-x) = -x , max(x,-x) = x > No you don't. You're adding the asiom -0 = 0. In fact you've postulated a special element 0 with the property -0 = 0. > It cannot be denied that there is some resemblance between the above > MinMax algebra - or whatever it should be called - and the rules of > common Boolean logic. There is an even closer relationship with the > rules of Fuzzy Logic. But these analogies are not quite unambiguous, > though I wish they were. The ambiguity is in the following choices: > It's looking much like a lattice ordered group. > - does an intersection / correspond with a Min or with a Max ? > - if then does a union / correspond with a Max or with a Min ? > (P(S), /, /), ie (P(S), intersection, uninn), is the lattice > of subsets S. Clearly A / B subset A,B subset A / B. > In particular, we have the formula for being approximately a subset: > X <= Y iff /_i /_j (x_i = y_j) where i=1,..,M and j=1,..,N > A subset B when for all a in A, a in B > iff for all a in A, some b in B with a = b > iff for all a in A, OR_(b in B) (a = b) > iff AND_(a in A) OR_(b in B) (a = b) Ok, got it this time. Have fun with infinite statements. > There's actually a logic that considers them. > Within the original posting, we have translated this formula into the > language of real numbers by equivalencing (x_i = y_j) with a distance, > equivalencing / with a product, and equivalencing / with a sum of > squares. It is noted that equivalencing / with a product is exactly > the other way around as is common in Fuzzy Logic. Which is troublesome. > The right formula in MinMax Technology is the other way around as well: > X <= Y iff max_i min_j ( |x_i - y_j| ) = residual (positive) > If X subset Y, then min_j |x_i - y_j| = 0. X subset Y iff max_(x in X) min_(y in Y) |x - y| = 0, > Left to right is nearly immediate as pointed out above. Right to left requires that max and min exist. When they don't, > you are not at liberty to use inf, sup in lieu of min, max. > X = {0}, Y = 1/N = { 1/n | n in N } for example. > Which is not in concordance with common Fuzzy Logic, as it would read: > X <= Y iff min_i max_j ( |x_i - y_j| ) = residual (positive) > Let X = { 0 }, Y = { 1 }. min_i max_j |x_i - y_j| = 1 > === Subject: Re: MinMax Technology > Let X be a finite set of M reals x_i : X = {x_1,x_2, .. ,x_i, .. ,x_M} > Let Y be a finite set of N reals y_j : Y = {y_1,y_2, .. ,y_j, .. ,y_N} > In order to determine whether X is a subset of Y, we could compare the > elements. Then X is a subset of Y, X <= Y , if and only if for each of > the x_i one of them is equal to y_j : > What if I first ordered each set from smallest to largest? > How would you do that, with e.g. elements in the Euclidean plane ? > You were discussing linear orders, not I. I was suggesting however, > that you extend your ponderings unto multi-dimensional logics. Uhm, yes. I've been sloppy (.. too often, it seems). My intention always has been to generalize this stuff to higher dimensions. Then |x_i - y_j| is the distance between _vectors_ (x_i e X) and (y_j e Y) in some space. I've done some excercises in 2-D. Sorry for being inaccurate here. > For what you ask, you'll want to linear order R^2 which can be > done, not with the product order, but the lexicographical order. You mean that I can accelerate the necessary comparisons in this manner? Can the following expression be evaluated faster with such an ordering: max_i min_j ( |x_i - y_j| ) ? > The rest of your comment certainly hold water, and I'll come back to it > You're welcome. You've hit upon one of my favorite interests, namely > order theory. It's exceeding more complex than what you'd expect from > the simplicity of (partial) orders. I've been revising my notes upon > ordering a partition of an ordered set and it's applications to > lattices, ordered groups and lattice ordered groups. Maybe I'll get it > together enough to present to the group. Again, I'll be wanting to know > why it too is not a counter example to William's Metatheorm: > Whatever math I dream up is already old hat. Han de Bruijn === Subject: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> Let X be a finite set of M reals x_i : X = {x_1,x_2, .. ,x_i, .. ,x_M} > Let Y be a finite set of N reals y_j : Y = {y_1,y_2, .. ,y_j, .. ,y_N} In order to determine whether X is a subset of Y, we could compare > the elements. Then X is a subset of Y, X <= Y , if and only if for > each of the x_i one of them is equal to y_j : > What if I first ordered each set from smallest to largest? > How would you do that, with e.g. elements in the Euclidean plane ? > For what you ask, you'll want to linear order R^2 which can be > done, not with the product order, but the lexicographical order. You mean that I can accelerate the necessary comparisons in this manner? > Can the following expression be evaluated faster with such an ordering: max_i min_j ( |x_i - y_j| ) ? > Ordering costs IIRC, are of order n^2. Arithmetical costs are significantly higher than comparison costs. How does the expression relate to set inclusion? Set inclusion is not symmetrical, yet your expression is. === Subject: Re: MinMax Technology > Let X be a finite set of M reals x_i : X = {x_1,x_2, .. ,x_i, .. ,x_M} > Let Y be a finite set of N reals y_j : Y = {y_1,y_2, .. ,y_j, .. ,y_N} > In order to determine whether X is a subset of Y, we could compare > the elements. Then X is a subset of Y, X <= Y , if and only if for > each of the x_i one of them is equal to y_j : > What if I first ordered each set from smallest to largest? > How would you do that, with e.g. elements in the Euclidean plane ? > For what you ask, you'll want to linear order R^2 which can be > done, not with the product order, but the lexicographical order. > You mean that I can accelerate the necessary comparisons in this manner? > Can the following expression be evaluated faster with such an ordering: > max_i min_j ( |x_i - y_j| ) ? > Ordering costs IIRC, are of order n^2. Arithmetical costs > are significantly higher than comparison costs. Yes. But how to _accomplish_ such a thing, in 2-D for example? > How does the expression relate to set inclusion? > Set inclusion is not symmetrical, yet your expression is. http://hdebruijn.soo.dto.tudelft.nl/jaar2008/index.htm Key reference, where it all started with: With MinMax technology as in the present posting: max_i min_j ( |x_i - y_j| ) But NO .. this expression is certainly _not_ symmetrical ! Han de Bruijn === Subject: Re: MinMax Technology <6df4e$48634746$82a1e228$19253@news1.tudelft.nl> <20080626030216.R16222@agora.rdrop.com> <71f87$4863ac8c$82a1e228$22526@news1.tudelft.nl> <20080627025724.I11436@agora.rdrop.com> <20080630020305.J31231@agora.rdrop.com> <8d8cf$4868bdf1$82a1e228$8116@news1.tudelft.nl > Let X be a finite set of M reals x_i : X = {x_1,x_2, .. ,x_i, .. ,x_M} > Let Y be a finite set of N reals y_j : Y = {y_1,y_2, .. ,y_j, .. ,y_N} > In order to determine whether X is a subset of Y, we could compare > the elements. Then X is a subset of Y, X <= Y , if and only if for > each of the x_i one of them is equal to y_j : > What if I first ordered each set from smallest to largest? > How would you do that, with e.g. elements in the Euclidean plane ? > For what you ask, you'll want to linear order R^2 which can be > done, not with the product order, but the lexicographical order. You mean that I can accelerate the necessary comparisons in this manner? > Can the following expression be evaluated faster with such an ordering: max_i min_j ( |x_i - y_j| ) ? > Ordering costs IIRC, are of order n^2. Arithmetical costs > are significantly higher than comparison costs. Yes. But how to _accomplish_ such a thing, in 2-D for example? > Are you talking algorithm? Likely the most efficient method for one time usage irregardless of what the subsets are is direct comparison OR_(a in A) AND_(b in B) (a = b) where of course A and B are finite because computers have no comprehension of anything infinite, using the neo-primitive counting method, one, two, many, overflow. Computational time appears to be o(n^2). === Subject: Uniform Continuity Sternberg, page 52. Let A be a subset of E^n. There exists a sequence of sets A_i (i>=1) and maps h_i (i>=1) such that A_0 is countable, A is included in Union of the A_r for r spanning between 0 and infinity, and for i>=1 1. h_i is a homeomorphism of the ball (B_r_i)^(m_i) into E^n with A_i included in h_i((B_r_i)^(m_i)) ((B_r)^m is the set of x in E^m such that ||x||= ||x-y|| 3. for any continuous function f vanishing on A there exist monotone functions b_i(r) that vanish for r->0 and such that |f(h_i(x))| < b_i(||x-y||) for all x, y in (B_r_i)^(m_i) with h_i(y) in A_i (and therefore f(h_i(y))=0). Now my problem is Sternberg claims that taking A_i = A / K_i (where K_i is a set of closed balls constituting a basis for E^n) and h_i to be the translation of the ball centered at zero to K_i it is clear that part (3.) is just uniform continuity. WHY IS THAT? P.S. How can a set of closed balls be a basis for E^n? Basis in what sense? I hope someone answers 'cause I'm in deep deep trouble. === Subject: Re: Uniform Continuity posting-account=JbgxqAoAAABtpHDW0Y3Rj06f1KAxfL77 Gecko/20080404 FIrefox/1.5 (Debian-2.0.0.14-2),gzip(gfe),gzip(gfe) > Sternberg, page 52. Let A be a subset of E^n. There exists a sequence of sets A_i (i>=1) and > maps h_i (i>=1) such that A_0 is countable, A is included in Union of the > A_r for r spanning between 0 and infinity, and for i>=1 1. h_i is a homeomorphism of the ball (B_r_i)^(m_i) into E^n with A_i > included in h_i((B_r_i)^(m_i)) ((B_r)^m is the set of x in E^m such that > ||x|| h_i. 2. ||h_r(x)-h_r(y)||>= ||x-y|| 3. for any continuous function f vanishing on A there exist monotone > functions b_i(r) that vanish for r->0 and such that |f(h_i(x))| < > b_i(||x-y||) for all x, y in (B_r_i)^(m_i) with h_i(y) in A_i (andsome > therefore f(h_i(y))=0). Now my problem is Sternberg claims that taking A_i = A / K_i (where K_i > is a set of closed balls constituting a basis for E^n) and h_i to be the > translation of the ball centered at zero to K_i it is clear that part (3.) > is just uniform continuity. For example, let b_i(t) = sup { |f(h_i(x))| : x,y in (B_r_i)^(m_i) with h_i(y) in A_i and ||x-y|| < t }. It is easy to see b_i is monontone (increasing t makes the sup'ed set larger) and the fact that b_i(t) -> 0 as t -> 0 follows from the uniform continuity of the function f(h_i(x)). (If you don't see this right away, it will become clear when you try to prove it.) > WHY IS THAT? P.S. > How can a set of closed balls be a basis for E^n? Basis in what sense? I think in the sense of being a basis for the topology on E^n. That is, every x in E^n has a neighborhood U which is contained in one of the K_i. A good example would be all closed balls with rational centers and rational radii. === Subject: Re: Uniform Continuity > For example, let b_i(t) = sup { |f(h_i(x))| : x,y in (B_r_i)^(m_i) with > h_i(y) in A_i and ||x-y|| < t }. It is easy to see b_i is monontone > (increasing t makes the sup'ed set larger) and the fact that b_i(t) -> 0 > as t -> 0 follows from the uniform continuity of the function f(h_i(x)). > (If you don't see this right away, it will become clear when you try to > prove it.) > Yes but all I know is that f is continuous, not uniformly continuous === Subject: New mathematics / physical sciences positions at http://jobs.phds.org, Jun 30, 2008 There are new job listings at http://jobs.phds.org -------------------------------------------------------------------- Title: Part-Time, Occasional Consulting for Thought Leaders in Energy Employer: Goldman Sachs - Vantage Marketplace Location: United States and Canada, United States Vantage is seeking knowledge leaders to join our world-class network of energy consultants from all disciplines in energy and/or vendors to the energy industry, with experience and knowledge in any of the following areas: * Exploration * Production... 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Full details: http://jobs.phds.org/job/9285/ben-gurion-university/post-doc-positions-in -------------------------------------------------------------------- Post your job (free!): http://jobs.phds.org/job/post PhDs.org: Science, Math, and Engineering Career Resources --------------------------------------------------------- * Job Listings: http://jobs.phds.org/ - Job board with hundreds of listings for Ph.D.s - Reach tens of thousands of Ph.D.s each month * Graduate School Rankings: http://graduate-school.phds.org/ - Comprehensive, customizable rankings of graduate programs * Career Resources: http://www.phds.org/ - Pointers to the best resources on the web for: + getting into graduate school + writing your dissertation + jobs for Ph.D.s in academia and industry * Engineering Science Weblog: http://blog.phds.org/ - Building better scientists and engineers === Subject: Geometry with radius.. Hello teacher~ A circle of radius 8.5 cm is circumscribed about a triangle with angles of 100, 34, and 46. Find the radius of the inscribed circle. --------------------------------------------------------- Sorry, I need your advice. Author's explanation) This problem was given to a group of ten high school mathematics teachers. Eight of them drew the problem incorrectly, and only half of them solved the problem within the 15 minute time span allowed. === Subject: Re: Geometry with radius.. > Hello teacher~ A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. Find the radius of the inscribed circle. --------------------------------------------------------- > Sorry, I need your advice. Author's explanation) > This problem was given to a group of ten high school mathematics teachers. > Eight of them drew the problem incorrectly, > and only half of them solved the problem within the 15 minute time span > allowed. a b c R = ------- = ------- = ------- sin A sin B sin C r = sqrt{(s-a)(s-b)(s-c)/s} = A/s 2.A = a.b.sin C = R^2 sin A sin B sin C = 2.s.r -- Michael Press === Subject: Re: Geometry with radius.. reply-type=response mina_world schrieb > Hello teacher~ A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. Find the radius of the inscribed circle. You can find the sides with the Law of Sines: a/sin(A) = b/sin(B) = c/sin(C) = 2*R (R = circumradius) When you know the sides, you can compute the inradius: r*sqrt((s-a)*(s-b)*(s-c)/s) Jutta === Subject: Re: Geometry with radius.. posting-account=33KaEgkAAAA9tz8WICNABjrkyMKXFbGS Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Hello teacher~ A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. (Author's explanation) > This problem was given to a group of ten high school mathematics teachers. > Eight of them drew the problem incorrectly, > and only half of them solved the problem within the 15 minute time span > allowed. One (student or teacher) needs to take a good look at in-radius and circum-radius formulas to understand inter-dependence of sides, R and r. After that, Time being let sides (a, b, c) be sin(100) , sin(40), sin(46) before scaling them up. Compute Area. Find out circum circle radius a b c / (4 Area). Find out linear scale factor( 8.5 / what you got )= SF. Compute in-radius from mathworld formula and scale it up( multiply by SF ). HTH === Subject: Re: Geometry with radius.. > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. > One (student or teacher) needs to take a good look at in-radius and > circum-radius formulas to understand inter-dependence of sides, > R and r. After that, [ snip a strange way of doing calculations ] > Compute in-radius from mathworld formula You don't need mathworld formula if you really did what you said. Or if you want to use mathworld formula 'from the book', just plug A,B,C angles and R into that formula, full stop. Rule for student : never apply a formula 'from a book' without understanding _why_ this formula, that is prooving it. So, 'after that', you should have _prooved_ mathworld formula ! (if you want doing all what you said is worth the effort) -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Geometry with radius.. posting-account=33KaEgkAAAA9tz8WICNABjrkyMKXFbGS Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. > One (student or teacher) needs to take a good look at in-radius and > circum-radius formulas to understand inter-dependence of sides, > R and r. > After that, [ snip a strange way of doing calculations ] > Compute in-radius from mathworld formula You don't need mathworld formula if you really did what you said. > Or if you want to use mathworld formula 'from the book', just plug > A,B,C angles and R into that formula, full stop. Rule for student : never apply a formula 'from a book' without > understanding _why_ this formula, that is prooving it. So, 'after that', you should have > prooved_ mathworld formula !(if you want doing all what you said is worth the effort) Agreed, but it is matter of knowing how much the student is already into. Rightly or wrongly anyway I assumed that formula for R ( a b c / 4 Area) is known, r(Area/s) is not so well known and could be looked up if familiarity had just started building up and required key relation r/4R = sin(A/2)sin(B/2)sin(C/2) not yet familiar at all, so may not readily use it and so on.(Roughly my own situation when at school!). Since angles are given, proportion of sides is fixed so it is a ready basis to start. 1 unit circum-diameter can be assumed, first get on to calculations and later on scale up ... is bit roundabout and yes perhaps not so convenient. What Jutta suggested is the best and direct under those circumstances. > -- > Philippe Ch., mail : chephip+n...@free.fr > site :http://mathafou.free.fr/ (recreational mathematics) Narasimham === Subject: Re: Geometry with radius.. > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. > One (student or teacher) needs to take a good look at in-radius and > circum-radius formulas > Compute in-radius from mathworld formula > You don't need mathworld formula if you really did what you said. > Rule for student : never apply a formula 'from a book' without > understanding [it] Agreed, but it is matter of knowing how much the student is already > into. Rightly or wrongly anyway I assumed that formula for R ( a b c / > 4 Area) is known, OK, that's the common start point : Law of sines : a/sin(A) = b/sin(B) = c/sin(C) = 2R = abc/(2*Area) [1] (afterall you NEED a relation with the angles, it's the only data !) > r(Area/s) is not so well known But however you NEED some kind of relation with inradius r ! You may use any one you like, or know. For instance as Jutta did : Area = s.r, with Heron, resulting into r = sqrt((s-a)(s-b)(s-c)/s), with s = (a+b+c)/2 But simply Area = s.r [2] is enough (and easy to proof/understand : Area of triangle is the sum of triangles through incenter, each one with altitude = r and base = one side. Just combining relations [1] and [2] results into : r = 2*R * sin(A)*sin(B)*sin(C) / (sin(A) + sin(B) + sin(C)) And you are finished. After that, a few trigonometric identities allow to simplify sin(A)*sin(B)*sin(C) / (sin(A) + sin(B) + sin(C)) into the mathworld formula (3) : > ... key relation r/4R = sin(A/2)sin(B/2)sin(C/2) not yet > familiar at all, Shure ! Not familiar to anybody at all. I just discovered it from this discussion ! > 1 unit circum-diameter can be assumed, > first get on to calculations and later on scale up ... is bit > roundabout and yes perhaps not so convenient. Shure, why not call R the circum-radius and do all the stuff with R instead of 1 ? No scaling is needed... And BTW, you can crosscheck your calculation through dimension equations, that is an area is a length^2, if you get a length^3, this shows you made a mistake, what you can't ever see if you get 1^3 without mentionning that this '1' is a length, while the '2' somewhere else is not and is just a dimensionless number. -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Geometry with radius.. > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. But however you NEED some kind of relation with inradius r ! > You may use any one you like, or know. For instance as Jutta did : Area = s.r, with Heron, resulting into > r = sqrt((s-a)(s-b)(s-c)/s), with s = (a+b+c)/2 > This is how I did it: 1. let A=46Á, B=34Á, C=100Á 2. construct Incenter I as intersection of bisectors A and B 3. project I onto AB = c with projections p, q 4. letting in-radius = 1: p = 1/tan(A/2), q = 1/tan(B/2) 5. construct point C with projections c1, c2, where c1+c2 = c (using angles A and B) 6. let h be the height of C above c = AB 7. As the triangle has C > 90Á the circum-center M lies below c = AB. call the distance m2. 8. coordinates of M with respect to origin A and x-axis c are (m1,-m2), where m1 = c/2 11. rescale so that R becomes 8.5. Time used: 15 minutes for the computation and another 3/4 hour to get it right, because I made two mistakes in the m2 computation. The first one was +h^2 instead of -h^2. The second was 1/h instead of 1/(2h). What a shame :-) Rainer === Subject: Re: Geometry with radius.. <6csp93F3hnpdjU1@mid.individual.net> posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. > But however you NEED some kind of relation with inradius r ! > You may use any one you like, or know. > For instance as Jutta did : Area = s.r, with Heron, resulting into > r = sqrt((s-a)(s-b)(s-c)/s), with s = (a+b+c)/2 This is how I did it: > 1. let A=46Á, B=34Á, C=100Á > 2. construct Incenter I as intersection of bisectors A and B > 3. project I onto AB = c with projections p, q > 4. letting in-radius = 1: p = 1/tan(A/2), q = 1/tan(B/2) > 5. construct point C with projections c1, c2, where c1+c2 = c > æ æ(using angles A and B) > 6. let h be the height of C above c = AB > 7. As the triangle has C > 90Á the circum-center M lies below > æ æc = AB. call the distance m2. > 8. coordinates of M with respect to origin A and x-axis c are > æ æ(m1,-m2), where m1 = c/2 > 11. rescale so that R becomes 8.5. Time used: 15 minutes for the computation and another 3/4 hour to > get it right, because I made two mistakes in the m2 computation. > The first one was +h^2 instead of -h^2. > The second was 1/h instead of 1/(2h). What a shame :-) Rainer ************************************************************** Nonsense. I don't think it is a shame. First, I've been teaching high school maths to some students here and there, and this problem is well beyond the usual, standard level of the highest high school maths track here in Israel, in spite of what is written in that book from where the story and the problem were taken. Second, the problem is not a geometry one but rather a trigonometry one. Third, unless one is given certain formulae and relations, it can take quite a while to come up with the solution of this problem. I know for sure that relations between incenter and circumcenter are not basic, standard thing in the geometry-trigonometry curriculum here, and unless one's taught and/or given those specific relations it could take a long time to evaluate all the things one need to solve the problem. Tonio === Subject: Re: Geometry with radius.. > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. Sorry, I need your advice. But however you NEED some kind of relation with inradius r ! > You may use any one you like, or know. > This is how I did it: > 1. let A=46Á, B=34Á, C=100Á > 2. construct Incenter I as intersection of bisectors A and B > 3. project I onto AB = c with projections p, q > 4. letting in-radius = 1: p = 1/tan(A/2), q = 1/tan(B/2) Hi Rainer, As I mentioned, you'd rather just call r the in-radius instead of 1 This makes easier cross-checking the calculations. > [...] > Time used: 15 minutes for the computation and another 3/4 hour to > get it right, because I made two mistakes > What a shame :-) Nonsense. I don't think it is a shame. My own calculations mistakes were in reducing 2*2*2/4 :-( First, [...] this problem is well beyond the usual, standard level of > the highest high school maths track [...] > I don't know exactly the corresponding school levels here and there. In France (in old programs) this kind of relations (the elementary ones, not that given in Mathworld) were tought at about age of 16/17. > Second, the problem is not a geometry one but rather a trigonometry > one. As soon as data is angle values, it is a trigonometry problem ! > Third, unless one is given certain formulae and relations, it can take > quite a while to come up with the solution of this problem. With a reasonable skill, we just need : area = base*altitude/2 sin, cos, tan definitions in a right triangle and properties of inscribed angles in a circle. But this requires a bit more than 15 minutes (mainly to find the right process). Knowing a few additional formulas just accelerates the calculations. > I know for sure that relations between incenter and circumcenter > are not basic Depends on which ones ! Basic relations : Area = abc/(4*R), Area = s*r and a = 2*R*sin(A) are enough here. There are a lot of weird relations which are not *tought*, but done as exercises. We usually never have to care of those. With just the above mentioned relations, we get the results in just a few lines. These are kind of elementary relations we see as soon as we deal with trigonometry and incircle. Jutta's method is also fine, but the r = sqrt(...) relation is not so well known (and even faster forgotten). For the sine law, one remembers easily the 1st part : a/sin(A) = b/sin(B) = c/sin(C) the rest (= 2R) is quickly forgotten. So we have (I had) to find it again from the basic circumcircle with inscribed angles properties, to get a = 2*R*sin(A). Relation Area = abc/4R is not so well known, but results at once from the previous one, so that we can forget it, and proove it again each time we need it, from a much usefull relation, hence less easily forgotten : Area = (1/2)*b*c*sin(A), with the above sin(A) = a/(2*R) So the amount of relations which are required to solve the given problem in less than 15 mins (including the effective numeric calculation !) is quite small, and these are only basic relations. After that, it is a question of skill in mixing these relations the right way (to stay in the 15 mins delay). -- Philippe Ch., mail : chephip+news@free.fr site : http://mathafou.free.fr/ (recreational mathematics) === Subject: Re: Geometry with radius.. > Hello teacher~ A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. Find the radius of the inscribed circle. --------------------------------------------------------- > Sorry, I need your advice. Author's explanation) > This problem was given to a group of ten high school mathematics > teachers. Eight of them drew the problem incorrectly, > and only half of them solved the problem within the 15 minute time span > allowed. See formula (3) at . David === Subject: Re: Geometry with radius.. > Hello teacher~ > A circle of radius 8.5 cm is circumscribed about a triangle > with angles of 100, 34, and 46. > Find the radius of the inscribed circle. > --------------------------------------------------------- > Sorry, I need your advice. > Author's explanation) > This problem was given to a group of ten high school mathematics > teachers. Eight of them drew the problem incorrectly, > and only half of them solved the problem within the 15 minute time span > allowed. See formula (3) at >http://mathworld.wolfram.com/Inradius.html>. If you need... http://board-2.blueweb.co.kr/user/math565/data/math/inscribed85.jpg === Subject: A question to number theorists: Does exist any direct proof that some number lesser than 1 is irrational? posting-account=McZ3aQkAAADz6LV-boDe1LcriRhf3lj3 Gecko/2008052906 Firefox/2.0.0.9;MEGAUPLOAD 1.0,gzip(gfe),gzip(gfe) > R matrix with elements i/j = primary rational number, i, j = > natural numbers 1 till n. > There are as many different rational numbers lesser than 1, as > there are natural numbers and primary rational number greater > than 1, together. > R matrix can be approximately divided into quarters: > 0-0.5; 0.5-1; 1-2; 2-n. > In the interval 1-2 are as many rational numbers as in the > interval 0.5-1. > In the interval 2-n are as many rational numbers as in the > interval 0-0.5. > Primary rational numbers greater than 1 are less dense than > in the range 0-1. > Between them thus exists enough place for irrational numbers. > Therefore my question is justified. > Fractions as (square root of 2)/2, e/3 or pi/4 are not accepted as > a proof. kunzmilan === Subject: Re: A question to number theorists: Does exist any direct proof that some number lesser than 1 is irrational? > Therefore my question is justified. > Fractions as (square root of 2)/2, e/3 or pi/4 are not accepted as > a proof. >kunzmilan Direct proof: .01001000100001000001... (continuing the obvious pattern) is irrational because its decimal representation does not repeat. --Lynn === Subject: Re: A question to number theorists: Does exist any direct proof that some number lesser than 1 is irrational? posting-account=H1y7YgoAAADzGQwbcYaL9UvwttgsjOjp AppleWebKit/525.18 (KHTML, like Gecko) Version/3.1.1 Safari/525.18,gzip(gfe),gzip(gfe) > Primary rational numbers ægreater than 1 are less dense than > in the range 0-1. Nonsense. > Between them thus exists enough place for irrational numbers. > Therefore my question is justified. > Fractions as (square root of 2)/2, e/3 or pi/4 are not accepted as > a proof. It is hard to imagine what you would accept as a proof. -- GM === Subject: Re: A question to number theorists: Does exist any direct proof that some number lesser than 1 is irrational? Primary rational numbers greater than 1 are less dense than > in the range 0-1. Nonsense. Between them thus exists enough place for irrational numbers. > Therefore my question is justified. > Fractions as (square root of 2)/2, e/3 or pi/4 are not accepted as > a proof. It is hard to imagine what you would accept as a proof. > -- > GM but 1/pi is o.k. according to his conditions. Ciao Karl === Subject: Re: Orientation of a compact manifold posting-account=R7AgUAoAAADVFAtIe36IBmgohoHjZsKW Gecko/20061201 Firefox/2.0.0.14 (Ubuntu-feisty),gzip(gfe),gzip(gfe) > On 30 Jun., 00:22, Mariano Su.87rez-Alvarez > On 29 Jun., 21:28, Mariano Su.87rez-Alvarez > an orientation of a n-dimensional topological manifold X is a suitable > choice of generators of H n(X,X-x) for every point x in X. > Is it possible to say that if X can be given the structure of a > simplicial complex (or more general: a delta complex which is defined > at Hatcher's book) and 'draw arrows' on the embedded simplices such > that everything fits together, then X is orientable in the above > sense? If this is true, does anybody know a proof? There is a > definition of a fundamental class of an orientated manifold which is a > suitable generator of the H n(X) homology group, too. If the above is > possible, can one see such a fundamental class in the orientated > simplicial complex at once? Perhaps with the example 'boundary of the > standard 3-simplex' which is homeomorphic to S^2? > As you probably know, it all depends on what you mean by `everything > fits together'. > Unless you make that precise, your question is meaningless. > -- m > I am sorry for being unprecise. By 'everything fits together' I mean > the following: > If you draw a standard n-simplex in the R^n, you can choose which of > the vertices is the first one, which is the second, etc. which is > symbolized by drawing arrows on the edged, e.g.*---->----* for the > standard 1 simplex. One can think of a simplicial complex as glued > from n-simplices in a special way. By 'everything fits together' I > mean choosing a numbering of the vertices of each embedded simplex > such that if two simplices intersect at another common simplex, the > two directions on this intersection simplex coincide. I have the > feeling that this is possible iff the simplicial complex is orientable > in the above algebraic sense. > For a 2-complex, if I understand you correctly, you want > to orient all 1-simplices in such a way that for no 2-simplex > in the complex you have an oriented cycle. > Take for example the triangulation of RP^2 shown in >http:www.math.jhu.edu/~jmb/note/rp2tri.pdf>. Can you rient the edges > so that non of the triangles become oriented cycles? > (Keep in mind that the edges are identified...) > -- m Yes, this is exactly what i mean. Perhaps it would have been better to > at first. The RP^2 is non-orientable and I actually can't find such an > orientation. So it seems to be right, but is it true in higher > dimensions, too? > S. Oh, I hope you have understood me in the correct way: *I* should have === Subject: even number posting-account=9mFhxAoAAADvYLsbTTuTch-3O-0ByiBY SIMBAR={C546AFFB-8196-4A15-9A40-D42F2869CD2F}),gzip(gfe),gzip(gfe) this is the question which I saw in a book :- can any even number greater than 2 be expressed as the product of primes ? the answer is given as : people just don't know === Subject: Re: even number this is the question which I saw in a book :- can any even number > greater than 2 be expressed as the product of primes ? the answer is given as : people just don't know I think you are referring to the famous Goldbach conjecture, which asks whether any even number greater than 2 can always be expressed as the SUM of two primes. People just don't know. Goldbach conjecture: can every even number greater than two always be expressed as the sum of two primes? Silverbach conjecture: can every sufficiently large odd number always be expressed as the sum of three primes? Bronzebach conjecture: can every even number always be expressed as the sum of two odd numbers? Woodenbach conjecture: can every odd number always be expressed as the sum of two even numbers? === Subject: Re: even number Stephen Montgomery-Smith Very Funny !!!! === Subject: Re: even number posting-account=G_G-iQoAAAB08LNQidt_LsMkopmIb4ZS Gecko/20060111 Firefox/1.5.0.1 Mnenhy/0.7.3.0,gzip(gfe),gzip(gfe) On Jun 30, 7:45 am, Stephen Montgomery-Smith > this is the question which I saw in a book :- can any even number > greater than 2 be expressed as the product of primes ? > the answer is given as : people just don't know I think you are referring to the famous Goldbach conjecture, which asks > whether any even number greater than 2 can always be expressed as the > SUM of two primes. People just don't know. Goldbach conjecture: can every even number greater than two always be > expressed as the sum of two primes? Silverbach conjecture: can every sufficiently large odd number always be > expressed as the sum of three primes? Bronzebach conjecture: can every even number always be expressed as the > sum of two odd numbers? Woodenbach conjecture: can every odd number always be expressed as the > sum of two even numbers? Leadbach conjecture: All primes are the sum of two smaller primes. Bill J === Subject: Re: even number posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; InfoPath.1),gzip(gfe),gzip(gfe) this is the question which I saw in a book :- can any even number > greater than 2 be expressed as the product of primes ? the answer is given as : people æjust don't know http://mathworld.wolfram.com/PrimeFactorization.html === Subject: Re: even number > this is the question which I saw in a book :- can any even number > greater than 2 be expressed as the product of primes ? the answer is given as : people just don't know The answer is: *any* number (even or odd, greater than 2 or not) can be expressed as a product of primes.This was proved a long, long time ago. What people do not know about even numbers greater than 2 is whether or not they can always be expressed as the _sum_ of _two_ prime numbers. The assertion yes, it is aways possible is known as Goldbach conjecture. Jose Carlos Santos === Subject: Re: even number <6cs26lF3hm4c4U1@mid.individual.net> posting-account=9mFhxAoAAADvYLsbTTuTch-3O-0ByiBY Gecko/20080418 Ubuntu/dapper-security Firefox/1.5.0.12eol,gzip(gfe),gzip(gfe) > this is the question which I saw in a book :- can any even number > greater than 2 be expressed as the product of primes ? > the answer is given as : people just don't know The answer is: *any* number (even or odd, greater than 2 or not) can be > expressed as a product of primes.This was proved a long, long time ago. What people do not know about even numbers greater than 2 is whether or > not they can always be expressed as the sum of two prime numbers. > The assertion yes, it is aways possible is known as Goldbach > conjecture. yeah you are correct, i quote the question wrongly === Subject: Sylvester minimal polynomial of x1 = Sqrt[2] + 3^(1/3)] m1[x]=__________________. minimal polynomial of x2 = 1/Sqrt[3^(1/3) + 5^(1/5)] m2[x]=__________________. Det[Sylvester matrix[m1[x - y], m2[y],y]]//Factor then get minimal polynomial of x1+x2 m[x]_______________________________. === Subject: Re: Sylvester > minimal polynomial of x1 = Sqrt[2] + 3^(1/3)] > m1[x]=__________________. x = sqr 2 + cbr 1/3 x - sqr 2 = cbr 1/3 x^3 - 3x^2 sqr 2 + 6x - 2.sqr 2 = 1/3 x^3 + 6x - (3x^2 + 2)sqr 2 = 1/3 2(3x^2 + 2)^2 = (x^3 - 6x - 1/3)^2 Simplify === Subject: Re: #14 when journalism strays from the truth, it no longer is journalism; new book: Internet Book Publishing #14 Archimedes Plutonium anagram - Mr. Meticulous Pinhead > --- quoting the Argus Leader story of Archimedes Plutonium --- > === Subject: Re: #17 fascinated by hecklers and their psychological imbalance; new book: Internet Book Publishing #17 Archimedes Plutonium anagram - Manure-mouth Disciple > === Subject: Re: #14 when journalism strays from the truth, it no longer is journalism; new book: Internet Book Publishing posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e InfoPath.1; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > --- quoting the Argus Leader story of Archimedes Plutonium --- > On the streets of Vermillion, people call him Archie. In Meckling, where he lives, children call him Bee Man as he walks > through town in a beekeeper's suit. But he keeps no bees. And online, where he is internationally famous, mathematicians and > scientists call him a kook, a crank and a nuisance. > --- end quoting --- Now I want to focus on this opening paragraph, not only because I > never wore a beekeeper's suit, but > I never owned one, and never touched one. I do have many mosquito net > jackets like the ones > fishermen wear in the wilds and creeks and streams to keep the > mosquitos at bay. I wear a mosquito > jacket over a T-shirt whenever I anticipate swarms while mowing or > working near wooded areas. > *************************************************************** Well, well: you've finally attained fame, though not glory...yet. And who cares what the children say? They see mosquito net jackets, and the lil' bastards think it is a beekeeper suit. They're stupid! Anyway, as they say: you're now internationally famous! Tonio Ps Save one autograph for me, please. === Subject: Re: #15 what is the journalistic practice of letting in other voices in a story?; new book: Internet Book Publishing posting-account=Rqa4sAoAAAC88UYanCtJRUF4S6TUauGA Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Is it fair to bring in Jesse Hughes who knows little to none of > physics on a discussion story of the Atom Totality? I have yet to find a thread that isn't improved by the presence of Jesse Hughes. Marshall === Subject: Re: #15 what is the journalistic practice of letting in other voices in a story?; new book: Internet Book Publishing #15 Archimedes Plutonium anagram - I am rectum unpolished > There must be some guidelines as to giving voice to others in a story === Subject: Re: Einstein s Lamentation That Special Relativity Was Not A Fundamental Theory > Right, but it's not an attempt to learn. What he's doing is > setting up an impossible standard for Greg, as if Greg is > wrong unless he can convince Ken. >As is the standard for many cranks here, as though physical >understanding were an entitlement rather than the result of hard work. > Can you imagine how demeaning it is to be called a crank by a self > righteous hypocrite like you? Obviously you're badly in need of some > elementary Canadian sensitivity training and anger management. Well, I'm not in the least angry, so I'm not sure what anger >management would offer in the present case. As for sensitivity >training, you may have a point. I invite you to illustrate the proper >Canadian way of educating someone like Seto that I have characterized >as a crank. I look forward to following your model. PD > Here's looking at me, Kid. A critical distinction between science and academics is that science concerns itself with physical issues such as relative motion studies in the case of relativity and academics are concerned with doctrinal exegesis, the definiition and significance of terminology and so on. This is what marks you as an academic rather than a scientist as you claim. Everyone around here discusses relativity as if analysis of the doctrines of Einstein were all that mattered when it is the relative motion studies and their resolution which matter regardless of what Einstein said or others opinions of what Einstein had to say on such subjects.And it's futile to discuss relativity in academic terms until there is a definite resolution of the underlying physical problems. ~v~~ === Subject: Indiscrete Thoughts, by Gian-Carlo Rota posting-account=5dWnDwkAAAAJX56s4Iiwi0owHzqwvK-v CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) Just got this, and I have to say this is the one of the most entertaining and books about mathematicians I have come across. His comments about people like Alonzo Church, Emil Artin, Solomon Lefschetz, and Stanislaw Ulam are amazingly perceptive and often downright funny. He really brings these people to life and shows them with all their human weaknesses. The writing is very lively and no-holds-barred. In one bio of Rota, it notes that Dr. Rota commented that when the book was first published one mathematician http://web.mit.edu/newsoffice/1999/rota.html Here is one review. http://www.maa.org/reviews/indiscthots.html The volume also includes the essay The Pernicious Influence of Mathematics Upon Philosophy, a highly opinionated piece that got Rota blacklisted in analytical philosophy circles. === Subject: Re: Indiscrete Thoughts, by Gian-Carlo Rota [Gian-Carlo Rota's] comments about people > like Alonzo Church, Emil Artin, Solomon Lefschetz, > and Stanislaw Ulam are amazingly perceptive How do you know? -- He is not here; but far away The noise of life begins again And ghastly thro' the drizzling rain On the bald street breaks the blank day. === Subject: Re: Indiscrete Thoughts, by Gian-Carlo Rota <486A3AD5.17071B83@tesco.net> posting-account=5dWnDwkAAAAJX56s4Iiwi0owHzqwvK-v NokiaE51-1/100.34.20; Profile/MIDP-2.0 Configuration/CLDC-1.1 ) AppleWebKit/413 (KHTML, like Gecko) Safari/413,gzip(gfe),gzip(gfe) On 1 hein.8a, 17:10, Frederick Williams [Gian-Carlo Rota's] comments about people > like Alonzo Church, Emil Artin, Solomon Lefschetz, > and Stanislaw Ulam are amazingly perceptive How do you know? That's a fair question and I suppose it's possible that Rota just had an over-active imagination and liked to make up stories about people that he knew. Still, the book was first published in 1997 and I'm not aware that anyone has called him a liar. Plus his anecdotes have the ring of truth to them. Doesn't prove a thing, I know, but his portraits are fascinating to read nonetheless. Rota' description of Stanislav Ulam, who was a personal friend, is the most detailed, but I was also intrigued to learn about Solomon Lefchetz's rudeness, Alonzo Church's odd machine like way of speaking, Emil Artin's dictatorial manners and cult-like following at Princeton and many other indiscrete observations. Even when Rota only says a little about someone, such as von Neumann, what he says is vivid and insightful, deepening my earlier understanding of his personality. === Subject: Re: Indiscrete Thoughts, by Gian-Carlo Rota > Just got this, and I have to say this is the one of the > most entertaining and books about mathematicians > I have come across. His comments about people > like Alonzo Church, Emil Artin, Solomon Lefschetz, > and Stanislaw Ulam are amazingly perceptive > and often downright funny. He really brings these > people to life and shows them with all their human > weaknesses. The writing is very lively and no-holds-barred. > In one bio of Rota, it notes that Dr. Rota commented > that when the book was first published one mathematician That is because up until then he never had a defensible excuse. Had he the courage of his convictions, he would have stopped talking long since. -- Michael Press === Subject: Re: Indiscrete Thoughts, by Gian-Carlo Rota posting-account=5dWnDwkAAAAJX56s4Iiwi0owHzqwvK-v CLR 1.1.4322; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > Just got this, and I have to say this is the one of the > most entertaining and books about mathematicians > I have come across. His comments about people > like Alonzo Church, Emil Artin, Solomon Lefschetz, > and Stanislaw Ulam are amazingly perceptive > and often downright funny. He really brings these > people to life and shows them with all their human > weaknesses. The writing is very lively and no-holds-barred. > In one bio of Rota, it notes that Dr. Rota commented > that when the book was first published one mathematician That is because up until then he never had a defensible excuse. > Had he the courage of his convictions, he would have stopped > talking long since. Apparently, Rota had a great interest in split personalities, in publically discussing the foibles and human weaknesses of great scientists. He took this interest to great lenghts, however, including discussing publically the way his friend Stanislaw Ulam's brain damage due to illness restricted Ulam's work in critical ways (Ulam was one of the two people who signed the patent application for the Hydrogen bomb; the other was Edward Teller). Here is an example from the book of the sort of anecdoates Rota recounts about Ulam: On rare occasions he felt overwhelmed by guilt at his inability to concentrate, which he viewed as avoidance of 'serious' work. He looked at me, his glacous eyes popping and twitching (they were the eyes of a medium, like Madama Blavatzky's), his mask about to come down, and asked, 'Isn't it true that I am a charlatan?' I proceeded to set his mind at rest by giving him, as a sedative, varied examples of flaming charlatans we both knew (with or without Nobel prizes). But soon his gnawing doubts would start over. He knew he would remain to the end a Yehudi Menuhin who never practised. (p. 79) It wasn't only fellow mathematicians who were offended and thought he was out of line. In the epilogue to Indiscrete Thoughts Rota's editor Fabrizio Palombi describes the reaction of Stanislaw Ulam's wife, Francoise, to these public revelations: What made him single out Stan Ulam's foibles for extensive description? Doesn't he know that one does not say think kind of thing about great men? He has had to bear the consequences of his actions. Mrs. Ulam made sure that he learned his lesson. She personally crossed out his name from the list of participants at every meeting that has taken place in the memory of Ulam. (p. 269) === Subject: Re: dedAnoe's Alchemistric.A posting-account=fNhuGwoAAAAcu_mWZdkQrPnHIPR_gQVH CLR 1.0.3705; .NET CLR 1.1.4322; Media Center PC 4.0),gzip(gfe),gzip(gfe) its just one picture === Subject: Re: #13 mistakes in the Archimedes Plutonium's story by Argus Leader; new book: Internet Book Publishing posting-account=JpxxPAgAAAAgwzQIYqn4j6syK-YhOmcF Gecko/20080201 Firefox/2.0.0.12,gzip(gfe),gzip(gfe) On Jun 30, 12:58 am, plutonium.ar Here is the website for this story: By the way, if you *do* wear bee suit, would you mind to post the picture of you in that suit? I think you'll look awfully cute, dressed in bee suit. If I were a bee, I couldn't resist. === Subject: Re: #13 mistakes in the Archimedes Plutonium's story by Argus Leader; new book: Internet Book Publishing #13 Anagram of Archimedes Plutonium Mister Eunuch Diploma === Subject: Re: #12 Archimedes Plutonium on front cover of Argus Leader newspaper story; new book: Internet Book Publishing > Well, it is out there, Archimedes Plutonium on the front page and full Never heard of it. I was on the cover of Wired magazine once - it was equally uncomplimentary. === Subject: Re: #12 Archimedes Plutonium on front cover of Argus Leader newspaper story; new book: Internet Book Publishing #12 anagram of Archimedes Plutonium Old-time U.S. Urine Champ === Subject: Re: #16 egregious newspaper reporter error homemade cape?; new book: Internet Book Publishing #16 Archimedes Plutonium anagram - Chromium Panties Duel === Subject: Primes will always exist between X and 2X I am sorry if I can not express this as good as some other will. First let us look at the fact we only need to test to the square root of any number to prove if it is prime. This is a bit of a hint. We can then build a Cartesian coordinate system using the set A={2,3,4,5,...} Notice 1 is not part of this set. We will make both x and y axis equal to this set. The results or product of x*y will be in the center of the set. Notice the only numbers missing from the result are that of prime numbers. If you first multiply Set A by 2 then 3 and so on you remove the multiples in of the set that can are not prime So we can then make the set B={2,3,4,5,...X} Set c={X+1,X+2,X+3,...2X} the cycles where even numbers are multiplied times the set only remove even values. The one where odd numbers are multiplied only remove odd numbers. However odd numbers that are a product of a prime are going to only repeatedly remove the same values already removed by that prime. So in perspective you may as well multiply times a set of primes say set D={2,3,5,7 squareroot(2X)} However you can even leave 2 out of that set because we are checking between x and 2 times X Each new cycle using an increasing prime removes more values from set C. However once you pass 3 and go to 5 it starts to repeat some of the values. such as 15 is in both 3 and 5 or more important are the ones in both eliminations between x and 2x. So you could then look at what each number eliminates as a separate set. such as 3 produces 3c ={6,9,12,15,18...} 5 produces 5c ={10,15,20,25,...} and so on with each prime. These sets over lap each other in several areas. Also notice each set also has no prime numbers The combining of all the sets is the means to eliminating all but the primes between X and 2X So for any value X >= 2 there will be primes between X and 2X Another way to look at this is each odd value removes a fraction of the numbers. The values below each cycle (int)(2X/3), (int)(2X/4), (int)(2X/5), are the ones that remain each time. The only ones that really matter to a greater extent are the (int)(2x/primes)the rest are repeating work already done. What this means is the that you are reducing fractions of fractions so can never get the entirety of all the numbers. Because 1 is not a part of the sets to do that or set A. We can then test this for any positive integer value greater than 2. You could even substitute J+1,J+2... for X if you choose to do so so long as J is a positive integer. Which you could look at the result set differently as {X...2X-3,2X-2,2X-1,2X} The following may not be all correct but the above is. As I said earlier there was a bit of a hint. primes affect a range that is X squared which is why you only need to check to the square root of a value to determine if it is prime. The slope of a squared function (X^2) I believe is the first derivative or 2X. === Subject: Re: Primes will always exist between X and 2X >especially if you can disprove the primes between X and 2X. X must be >from set {2,3,4,5....} Nobody's going to disprove that there is always a prime between X and 2X because it's true. But what's below is simply not a proof. You say > The combining of all the sets is the means to eliminating all but the primes > between X and 2X So for any value X >= 2 there will be primes between X and 2X How do you _know_ that it can't happen that _all_ the numbers between X and 2X have been eliminated? <19430948.1214851900541.JavaMail.jakarta@nitrogen.mathforum.org>, > I am sorry if I can not express this as good as some other will. First let us look at the fact we only need to test to the square root of any > number to prove if it is prime. > This is a bit of a hint. We can then build a Cartesian coordinate system using the set A={2,3,4,5,...} > Notice 1 is not part of this set. > We will make both x and y axis equal to this set. > The results or product of x*y will be in the center of the set. Notice the > only numbers missing from the result are that of prime numbers. If you first multiply Set A by 2 then 3 and so on you remove the multiples in > of the set that can are not prime So we can then make the set B={2,3,4,5,...X} > Set c={X+1,X+2,X+3,...2X} the cycles where even numbers are multiplied times the set only remove even > values. > The one where odd numbers are multiplied only remove odd numbers. However odd > numbers that are a product of a prime are going to only repeatedly remove the > same values already removed by that prime. > So in perspective you may as well multiply times a set of primes say set > D={2,3,5,7 squareroot(2X)} However you can even leave 2 out of that set > because we are checking between x and 2 times X Each new cycle using an increasing prime removes more values from set C. > However once you pass 3 and go to 5 it starts to repeat some of the values. > such as 15 is in both 3 and 5 or more important are the ones in both > eliminations between x and 2x. So you could then look at what each number eliminates as a separate set. > such as 3 produces 3c ={6,9,12,15,18...} > 5 produces 5c ={10,15,20,25,...} > and so on with each prime. > These sets over lap each other in several areas. > Also notice each set also has no prime numbers > The combining of all the sets is the means to eliminating all but the primes > between X and 2X So for any value X >= 2 there will be primes between X and 2X Another way to look at this is each odd value removes a fraction of the > numbers. > The values below each cycle (int)(2X/3), (int)(2X/4), (int)(2X/5), are the > ones that remain each time. The only ones that really matter to a greater > extent are the (int)(2x/primes)the rest are repeating work already done. What this means is the that you are reducing fractions of fractions so can > never get the entirety of all the numbers. Because 1 is not a part of the > sets to do that or set A. We can then test this for any positive integer value greater than 2. You > could even substitute J+1,J+2... for X if you choose to do so so long as J is > a positive integer. Which you could look at the result set differently as {X...2X-3,2X-2,2X-1,2X} The following may not be all correct but the above is. > As I said earlier there was a bit of a hint. primes affect a range that is X > squared which is why you only need to check to the square root of a value to > determine if it is prime. The slope of a squared function (X^2) I believe is > the first derivative or 2X. you can disprove the primes between X and 2X. X must be from set > {2,3,4,5....} -- David C. Ullrich === Subject: Re: Primes will always exist between X and 2X posting-account=suWj4AkAAADE1IvGmj55Nmq3f98qb17e SIMBAR Enabled; SIMBAR={70306B22-CB8C-4d52-BFF4-18424E217075}; MathPlayer 2.10b; .NET CLR 2.0.50727),gzip(gfe),gzip(gfe) > I am sorry if I can not express this as good as some other will. First let us look at the fact we only need to test to the square root of any number to prove if it is prime. > This is a bit of a hint. We can then build a Cartesian coordinate system using the set A={2,3,4,5,...} Notice 1 is not part of this set. > We will make both x and y axis equal to this set. > The results or product of x*y will be in the center of the set. Notice the only numbers missing from the result are that of prime numbers. If you first multiply Set A by 2 then 3 and so on you remove the multiples in of the set that can are not prime So we can then make the set B={2,3,4,5,...X} > Set c={X+1,X+2,X+3,...2X} the cycles where even numbers are multiplied times the set only remove even values. > The one where odd numbers are multiplied only remove odd numbers. However odd numbers that are a product of a prime are going to only repeatedly remove the same values already removed by that prime. > So in perspective you may as well multiply times a set of primes say set D={2,3,5,7 squareroot(2X)} However you can even leave 2 out of that set because we are checking between x and 2 times X Each new cycle using an increasing prime removes more values from set C. However once you pass 3 and go to 5 it starts to repeat some of the values. such as 15 is in both 3 and 5 or more important are the ones in both eliminations between x and 2x. So you could then look at what each number eliminates as a separate set. > such as 3 produces 3c ={6,9,12,15,18...} > 5 produces 5c ={10,15,20,25,...} > and so on with each prime. > These sets over lap each other in several areas. > Also notice each set also has no prime numbers > The combining of all the sets is the means to eliminating all but the primes between X and 2X So for any value X >= 2 there will be primes between X and 2X Another way to look at this is each odd value removes a fraction of the numbers. > The values below each cycle (int)(2X/3), (int)(2X/4), (int)(2X/5), are the ones that remain each time. The only ones that really matter to a greater extent are the (int)(2x/primes)the rest are repeating work already done. What this means is the that you are reducing fractions of fractions so can never get the entirety of all the numbers. Because 1 is not a part of the sets to do that or set A. We can then test this for any positive integer value greater than 2. You could even substitute J+1,J+2... for X if you choose to do so so long as J is a positive integer. Which you could look at the result set differently as {X...2X-3,2X-2,2X-1,2X} The following may not be all correct but the above is. > As I said earlier there was a bit of a hint. primes affect a range that is X squared which is why you only need to check to the square root of a value to determine if it is prime. The slope of a squared function (X^2) I believe is the first derivative or 2X. > *************************************************** Google or yahoo Bertrand's Postulate. Tonio === Subject: Every Chess Configuration Can we write a program which will eventually display ALL possible chess games? Yes, you heard right. ALL possible chess games. You don't think so? Think again. Here's a short analysis with a Maple worksheet which does exactly that: http://ioannis.virtualcomposer2000.com/math/EveryChess.html idea. -- I.N. Galidakis === Subject: Re: Every Chess Configuration posting-account=kxPkPAoAAACjJi8w0gL9bnyznPzdw9HW Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Can we write a program which will eventually display ALL possible chess games? Yes, you heard right. ALL possible chess games. You don't think so? Think again. Here's a short analysis with a Maple worksheet which does exactly that: http://ioannis.virtualcomposer2000.com/math/EveryChess.html idea. > -- > I.N. Galidakis Put it this way how big opening book (all possible moves) would you be able to store on a 500 GB drive, if we do not allow drawback moves. How many moves ahead would we be able to store uncompressed? 15-20 or maybe 30? J === Subject: Re: Every Chess Configuration posting-account=kxPkPAoAAACjJi8w0gL9bnyznPzdw9HW Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Can we write a program which will eventually display ALL possible chess games? Yes, you heard right. ALL possible chess games. You don't think so? Think again. Here's a short analysis with a Maple worksheet which does exactly that: http://ioannis.virtualcomposer2000.com/math/EveryChess.html idea. > -- > I.N. Galidakis Well idea isn't that bad since you only need to examine 57 to 70 moves. Average number of moves per game: 57.63 black+white What would be interesting to know is how many possible configurations is possible after. 5,10,15... moves etc without any drawback moves. Those number would really be interesting to know. J === Subject: Re: Every Chess Configuration >Can we write a program which will eventually display ALL possible chess games? Yes, you heard right. ALL possible chess games. You don't think so? Think again. Here's a short analysis with a Maple worksheet which does exactly that: http://ioannis.virtualcomposer2000.com/math/EveryChess.html idea. See http://members.iinet.net.au/~ray/Chessgames.htm ray@iinet.com.au www.iinet.com.au/~ray http://en.wikipedia.org/wiki/On_Liberty The preventive function of government, however, is far more liable to be abused, to the prejudice of liberty, than the punitory function; for there is hardly any part of the legitimate freedom of action of a human being which would not admit of being represented, and fairly too, as increasing the facilities for some form or other of delinquency. J S Mill === Subject: Re: Every Chess Configuration Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Can we write a program which will eventually display ALL possible chess games? This is not very difficult, just impractical due to the large number. -- Richard -- Please remember to mention me / in tapes you leave behind. === Subject: Re: Every Chess Configuration posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI SLCC1; .NET CLR 2.0.50727; .NET CLR 1.1.4322; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) >Can we write a program which will eventually display ALL possible chess games? This is not very difficult, just impractical due to the large number. Using the calculation from here: http://www.chesscafe.com/text/bruce23.pdf {which is the most conservative estimate of 'the longest chess game possible' that I have seen} We have 5,898 moves * 2 plies/move = 11,796 plies. The estimate for the average branching factor of chess is 35. 35^(11796) = 6.71e+18213 which is a fairly substantial number, I would say. === Subject: Re: Every Chess Configuration Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) >Can we write a program which will eventually display ALL possible chess games? >This is not very difficult, just impractical due to the large number. Still, it does take thought and skill to enumerate all the possible chess games using only a fixed practical amount of program memory and disk space as working memory. -- I buy more from my grocer than he buys from me, and I bet it's the same with you and your grocer. That means we have a trade deficit with our grocers. Does our perpetual grocer trade deficit portend doom? -- Walter Williams === Subject: Re: Every Chess Configuration > This is not very difficult, just impractical due to the large number. Still, it does take thought and skill to enumerate all the possible > chess games using only a fixed practical amount of program memory > and disk space as working memory. And how would that be? #include #include #include #include const char *pieces[] = { BW, BW, BW, BW, BW, BW, BW, BW, TW, TW, SW, SW, LW, LW, KW, DW, BS, BS, BS, BS, BS, BS, BS, BS, TS, TS, SS, SS, LS, LS, KS, DS }; #define NUMPCS (sizeof(pieces) / sizeof(const char *)) void eval_pos(const unsigned char *pos) { int x, y; printf(+---+---+---+---+---+---+n); for (x = 0; x < 8; x++) { printf(|); for (y = 0; y < 8; y++) { if (pos[8 * y + x] == 0) printf( |); else printf(%s|, pieces[pos[8 * y + x]]); } printf(n); printf(+---+---+---+---+---+---+n); } printf(n); } int main(int argc, char **argv) { int i; unsigned char pos[64]; memset(pos, 0, sizeof(pos)); do { eval_pos(pos); for (i = 0; i < sizeof(pos); i++) { pos[i]++; if (pos[i] == NUMPCS) { pos[i] = 0; } else { break; } } } while (i != sizeof(pos)); return 0; } This program also emits unreachable positions and is a pile of dogcrap. However, it was not hard to write. Nor were memory constraints of any relevance, as you can easily see. Johannes -- Wer etwas kritisiert muss es noch lange nicht selber besser k.9annen. Es reicht zu wissen, da¤ andere es besser k.9annen und andere es auch besser machen um einen Vergleich zu bringen. - Wolfgang Gerber in de.sci.electronics <47fa8447$0$11545$9b622d9e@news.freenet.de> === Subject: Re: Every Chess Configuration Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) > This is not very difficult, just impractical due to the large number. > Still, it does take thought and skill to enumerate all the possible > chess games using only a fixed practical amount of program memory > and disk space as working memory. >And how would that be? >This program also emits unreachable positions A chess game is an ordered sequence of legal chess moves (that's -legal-, not -good-); the program you posted doesn't even enumerate all the possible legal chess board positions let alone the chess -games-. -- Ignorance has been our king... he sits unchallenged on the throne of Man. His dynasty is age-old. His right to rule is now considered legitimate. Past sages have affirmed it. They did nothing to unseat him. -- Walter M Miller, Jr === Subject: Re: Every Chess Configuration > This program also emits unreachable positions A chess game is an ordered sequence of legal chess moves > (that's -legal-, not -good-); the program you posted > doesn't even enumerate all the possible legal chess board positions > let alone the chess -games-. Yes, it does. Johannes -- Wer etwas kritisiert muss es noch lange nicht selber besser k.9annen. Es reicht zu wissen, da¤ andere es besser k.9annen und andere es auch besser machen um einen Vergleich zu bringen. - Wolfgang Gerber in de.sci.electronics <47fa8447$0$11545$9b622d9e@news.freenet.de> === Subject: Re: Every Chess Configuration Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) > This program also emits unreachable positions > A chess game is an ordered sequence of legal chess moves > (that's -legal-, not -good-); the program you posted > doesn't even enumerate all the possible legal chess board positions > let alone the chess -games-. >Yes, it does. Yes it does what? Enumerate all the possible legal chess games? Enumerate all the possible legal chess board positions? By definition, an enumeration of something is a list of all the valid states that something can be in, with no invalid states included in the list. Your own statement is that the program emits unreachable positions; is that statement incorrect and in fact the program does -not- emit unreachable positions, or is the statement correct and the program -does- emit unreachable positions? If it emits even *one* unreachable position, then it is not an enumeration of legal chess board positions. Not if we mean legal in the sense of according to the rules of chess rather than in the sense of criminal law. -- Beware of bugs in the above code; I have only proved it correct, not tried it. -- Donald Knuth === Subject: Re: Every Chess Configuration > Yes it does what? Enumerate all the possible legal chess games? > Enumerate all the possible legal chess board positions? Games I never talked about, simply positions. This is what the OPs program is doing, too. > By definition, an enumeration of something is a list of all > the valid states that something can be in, with no invalid states > included in the list. Alright, by that definition my program does not output an enumeration. It outputs chess positions among which are all possible valid positions. Although the OPs program does not output as much bullcrap as mine does, to be fair you have to also say that it is *not* an enumeration: There are valid positions in which more than two rooks, queens, bishops or knights may appear, as pawns might get promoted. This however requires that at least one pawn is taken out of the board (the one getting promoted). It also makes the enumeration of positions *way* more difficult and actually makes it probably easier to enumerate games to get the positions, as this judging which is a legal position is hardly trivial. Johannes -- Wer etwas kritisiert muss es noch lange nicht selber besser k.9annen. Es reicht zu wissen, da¤ andere es besser k.9annen und andere es auch besser machen um einen Vergleich zu bringen. - Wolfgang Gerber in de.sci.electronics <47fa8447$0$11545$9b622d9e@news.freenet.de> === Subject: Re: Every Chess Configuration Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) > Yes it does what? Enumerate all the possible legal chess games? > Enumerate all the possible legal chess board positions? Games I never talked about, simply positions. This is what the OPs >program is doing, too. >Can we write a program which will eventually display ALL possible chess games? Regardless of what the OP's program actually -does-, the goal expressed by the OP was in terms of *games*, not *positions*. And although enumerating all possible *games* might merely take a long time and be boring code of no particular skill in a naive implementation, writing an implementation that managed to enumerate all the possible *games* in a fixed and reasonable amount of memory and temporary disk space would take some skill. I did not look at the original program to determine whether that skill had been displayed in it: I did but point out that there is a reasonable constraint on the enumeration algorithm (i.e., that it actually be able to finish on a real system, given enough time) would turn the problem from a low-skill one to one of higher skill and moderate interest. -- Tired minds don't plan well. Sleep first, plan later. -- Walter Reisch === Subject: Re: Every Chess Configuration posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI 1.1.4322; .NET CLR 1.0.3705; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Jul 2, 9:33æam, rober...@ibd.nrc-cnrc.gc.ca (Walter Roberson) > Yes it does what? Enumerate all the possible legal chess games? > Enumerate all the possible legal chess board positions? >Games I never talked about, simply positions. This is what the OPs >program is doing, too. >Can we write a program which will eventually display ALL possible chess games? Regardless of what the OP's program actually -does-, the > goal expressed by the OP was in terms of *games*, not *positions*. > And although enumerating all possible *games* might merely take > a long time and be boring code of no particular skill in a naive > implementation, writing an implementation that managed to enumerate > all the possible *games* in a fixed and reasonable amount > of memory and temporary disk space would take some skill. I did not look at the original program to determine whether that > skill had been displayed in it: I did but point out that there is > a reasonable constraint on the enumeration algorithm (i.e., > that it actually be able to finish on a real system, given > enough time) would turn the problem from a low-skill one to one > of higher skill and moderate interest. Castling rights, half move clock, and e.p. status also change the nature of a position. A position with exactly the same piece placement on the board but a different half move clock value is really a different position. In addition, a position may be two moves from checkmate but actually be a draw because of the three move repetition rule, which can only be discerned by the actual game state. === Subject: Re: Every Chess Configuration Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >Can we write a program which will eventually display ALL possible chess games? >This is not very difficult, just impractical due to the large number. >Still, it does take thought and skill to enumerate all the possible >chess games using only a fixed practical amount of program memory >and disk space as working memory. On the other hand, since there is no propspect of the program finishing, you needn't waste any effort in optimising it for speed. -- Richard -- Please remember to mention me / in tapes you leave behind. === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess games? Sure. It's just totally pointless. Now if you could in fact enumerate all positions, remove the illegal ones, tell us which moves do get you there and then determine whether black or white has an advantage in *reasonable* time, I'll be impressed. Being generous, resonable time is, say, 10 years? Now I'll start writing a program which determines all possible screen contents of a 1600x1200 screen with 24 bits of color. Yay. What fun. Johannes -- Wer etwas kritisiert muss es noch lange nicht selber besser k.9annen. Es reicht zu wissen, da¤ andere es besser k.9annen und andere es auch besser machen um einen Vergleich zu bringen. - Wolfgang Gerber in de.sci.electronics <47fa8447$0$11545$9b622d9e@news.freenet.de> === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess > games? Sure. It's just totally pointless. Now if you could in fact enumerate all positions, remove the illegal > ones, tell us which moves do get you there and then determine whether > black or white has an advantage in *reasonable* time, I'll be impressed. towards impressing you. How could I EVER forget your existence? This must be a momentary memory slip on my part. I got side-tracked with this project. Again my apologies for not impressing you. I'll do my best next time. If you have any projects and work for me to do that may impress you when I am done with them, please feel free to email me! Aei gamisou skato-germanara > Johannes -- I.N. Galidakis === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess > games? > Sure. It's just totally pointless. > Now if you could in fact enumerate all positions, remove the illegal > ones, tell us which moves do get you there and then determine whether > black or white has an advantage in *reasonable* time, I'll be impressed. towards impressing you. How could I EVER forget your existence? This must be a > momentary memory slip on my part. I got side-tracked with this project. Again my apologies for not impressing you. I'll do my best next time. Haha, neat. I like getting impressed. And I like the part where you get all pissed and stuff. You know it's like saying: Do you think its possible to find the largest prime number, one no one has found before? Don't think it's greatest prime there ever was. Only thing is it takes about 1.44E890348 years. Is that impressive? No. Now if I said: a split second, its 2^78490734089783 - 1, *that* would impress people. > If you have any projects and work for me to do that may impress you when I am > done with them, please feel free to email me! Sure. I've found the largest prime number so far. Its 2^78490734089783 - 1. > Aei gamisou skato-germanara Wennde meinst Johannes -- Wer etwas kritisiert muss es noch lange nicht selber besser k.9annen. Es reicht zu wissen, da¤ andere es besser k.9annen und andere es auch besser machen um einen Vergleich zu bringen. - Wolfgang Gerber in de.sci.electronics <47fa8447$0$11545$9b622d9e@news.freenet.de> === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess > games? > Sure. It's just totally pointless. Now if you could in fact enumerate all positions, remove the illegal > ones, tell us which moves do get you there and then determine whether > black or white has an advantage in *reasonable* time, I'll be impressed. > towards impressing you. How could I EVER forget your existence? This must be > a momentary memory slip on my part. I got side-tracked with this project. > Again my apologies for not impressing you. I'll do my best next time. Haha, neat. I like getting impressed. You do, eh? Let me tell you *I* like: I like people who apply the law If you can't say anything NICE, don't say anything at ALL, as a response to a desire to SHARE with GOOD INTENT. I like people who don't repeat OBVIOUS stuff (about chess engines) I and everybody else knows, as a response to desire to SHARE with GOOD INTENT. I like people who don't repeat A SECOND TIME obvious stuff I and everybody knows (slightly adapted for prime numbers), as a response to a desire to SHARE with GOOD INTENT. I like people who don't have a SUPER INFLATED EGO and are not just looking for an excuse to proclaim ME TOO! ME TOO!, pointing out the obvious AGAIN as a response to SHARE with GOOD INTENT. > And I like the part where you get > all pissed and stuff. Yeah, it occasionally happens when a colossal moron shows up. I am particularly annoyed by ungratefulness as a RESPONSE to GOOD INTENT. The original post is obviously a trivially pointless exercise, since the times involved are huge. If computing skills, you've got the wrong idea there, somewhere. The IDEA is what's NEAT! I find this NEAT idea, which a friend sent and after working on it happily for 4 hours and having a GOOD TIME adapting it to chess, here I come with a DESIRE to SHARE with everybody on the newsgroup. This is like two kids playing, with one kid approaching the second kid and saying: Here, I have found this new toy. Would you like to have it and play with it, TOO? and with the second kid responding: Your toy is BULLCRAP! Even if the first kid's toy is INDEED bullcrap, can't you just keep your mouth SHUT? NOOOOO! I guess you can't resist your primitive urge to proclaim I can do that TOO! Me TOO! LOOK at ME, people! ME TOO. I mean, how OLD are you? 5? Do you GET it or do I have to call ing Freud here to spell it for you using LaTeX? > You know it's like saying: Do you think its possible to find the > largest prime number, one no one has found before? Don't think it's > greatest prime there ever was. Only thing is it takes about 1.44E890348 > years. Is that impressive? No. Now if I said: a split second, its 2^78490734089783 - 1, *that* would impress people. > If you have any projects and work for me to do that may impress you when I am > done with them, please feel free to email me! Sure. I've found the largest prime number so far. Its > 2^78490734089783 - 1. Holy Cow! Check it out! After the most sarcastic response one could have, this guy STILL doesn't get it. You are a bigger LOSER and more BORING than even *I* imagine you to be. > Aei gamisou skato-germanara Wennde meinst > Johannes -- I.N. Galidakis === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess games? Sure. It's just totally pointless. Now if you could in fact enumerate all positions, remove the illegal > ones, tell us which moves do get you there and then determine whether > black or white has an advantage in *reasonable* time, I'll be > impressed. Being generous, resonable time is, say, 10 years? Now I'll start writing a program which determines all possible screen > contents of a 1600x1200 screen with 24 bits of color. Yay. What fun. I did that, and ran it to completion. To prove that I did it, I can provide the brightest, darkest, and mean screen configurations encountered during the search. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Every Chess Configuration posting-account=b4rQkQoAAADqqftYxNCieauSttMs6_zU 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > Can we write a program which will eventually display ALL possible chess games? Yes, you heard right. ALL possible chess games. You don't think so? Think again. Here's a short analysis with a Maple worksheet which does exactly that: http://ioannis.virtualcomposer2000.com/math/EveryChess.html idea. > -- > I.N. Galidakis Cool. When will you be posting the output? -scattered === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess > games? > Yes, you heard right. ALL possible chess games. You don't think so? Think > again. > Here's a short analysis with a Maple worksheet which does exactly that: > http://ioannis.virtualcomposer2000.com/math/EveryChess.html > the idea. > -- > I.N. Galidakis Cool. When will you be posting the output? It's on its way. I've already started printing all positions. At my current printing rate of 480076 positions/sec, I expect to be done in around 1.294965e+58 years. I'll post a quick note to sci.math when I'm done. Now if you'll excuse me, I've got to go to Kinko's and order some more A4 paper... ;o) > -scattered -- I.N. Galidakis === Subject: Re: Every Chess Configuration <1214864145.177747@athprx04> posting-account=kxPkPAoAAACjJi8w0gL9bnyznPzdw9HW Mozilla/4.0 (compatible; MSIE 6.0; Windows NT 5.1; SV1) ; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 1.1.4322; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > Can we write a program which will eventually display ALL possible chess > games? > Yes, you heard right. ALL possible chess games. You don't think so? Think > again. > Here's a short analysis with a Maple worksheet which does exactly that: >http://ioannis.virtualcomposer2000.com/math/EveryChess.html > the idea. > -- > I.N. Galidakis > Cool. When will you be posting the output? It's on its way. I've already started printing all positions. At my current printing rate of 480076 positions/sec, I expect to be done in > around 1.294965e+58 years. I'll post a quick note to sci.math when I'm done. Now if you'll excuse me, I've got to go to Kinko's and order some more A4 > paper... ;o) > -scattered -- > I.N. Galidakis- D.9alj citerad text - - Visa citerad text - Well if you started what move are you on, that is more interesting. Do you allow immediate drawback of pieces that could be a really long games. Do you have any rules for moving back pieces in your program? J === Subject: Re: Every Chess Configuration > Can we write a program which will eventually display ALL possible chess > games? > Yes, you heard right. ALL possible chess games. You don't think so? Think > again. > Here's a short analysis with a Maple worksheet which does exactly that: > http://ioannis.virtualcomposer2000.com/math/EveryChess.html > the idea. > -- > I.N. Galidakis > Cool. When will you be posting the output? > It's on its way. I've already started printing all positions. > At my current printing rate of 480076 positions/sec, I expect to be done in > around 1.294965e+58 years. > I'll post a quick note to sci.math when I'm done. > Now if you'll excuse me, I've got to go to Kinko's and order some more A4 > paper... ;o) > -scattered > -- > I.N. Galidakis- D.9alj citerad text - > - Visa citerad text - Well if you started what move are you on, that is more interesting. Do > you allow immediate drawback of pieces that could be a really long > games. Do you have any rules for moving back pieces in your program? J I was trying to get an idea of the complexity of solving chess in my post. One idea is to list all positions (including a parameter to account for the 50 moves rule) that are checkmate for white, and to backtrack by 1 ply, leading to 1 ply mate by white. Then backtrack by 1 Black move, and only keep positions which in 1 white move, lead to mate in 1 ply by White. Then, backtrack by 1 White ply and keep 3-ply forced mates by White only. We're looking for a forced checkmate by White... If at some point in the back-tracking we reach the configuration before White's first move (initial set-up), then we've discovered a winning strategy for White. If none is found, repeat with checkmate positions by Black and backtrack, trying to reach the initial board configuration. We need to put 32 pieces in 64 squares. 64 (White king) 63 (black king) 62 (white queen) 61 (black queen) 60*59/2 (2 white rooks) 58*57/2 (2 black rooks) 56*55/2 (2 white bishops) 54*53/2 (2 black bishops) 52*51/2 (2 white knights) 50*49/2 (2 Black knights) 48*... * 41/(8!) (8 white pawns) 40*... *33/ (8!) (place 8 Black paws) Answer : product = (64!)/((32!) * (2!)^6 * (8!)^2) ~ = 4.6 * 10^42 . We should remove impossible to reach positions, and add positions including pawn promotions and positions with less than 32 pieces. Also, a minor contribution is the castling state for Black/White and the 50-move counter. Actually, one could start with a subset of the White checkmates, and if backtracking gives a forced mate by White, it would be solved. My intuition would say that with both sides playing their best, in the end-game captures by White will come in bursts W x, Black x, W x, Black x, W x, Black x -> 6 pieces removed or W x, Black x -> 2 pieces removed and between these capture/recaptures, there could be many moves without a capture, in the end game at least. So 150-move end-games seem plausible, unless there's a forced mate by one side in less than 150 moves. Even if one hypothesizes a possible forced mate by White from 1. e4 (or whatever), it could be 80 moves or 200 moves. The point is that backtracking by 200 moves is much much harder than backtracking by 80 moves. And if neither side can force a win, it seems to me that would be case where chess solving is the most intractable ... David Bernier === Subject: Re: Every Chess Configuration Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) >and between these capture/recaptures, there could be many moves >without a capture, in the end game at least. So 150-move >end-games seem plausible, unless there's a forced mate >by one side in less than 150 moves. Wikipedia indicates, The longest tournament chess game ever to be played under modern time rules was Nikolic'-Arsovic', Belgrade 1989, which lasted for 20 hours and 15 minutes, ending in a 269-move draw.[1][2] At the time this game was played, FIDE had modified the fifty move rule to allow 100 moves to be played without a piece being captured in a rook and bishop versus rook endgame, the situation in Nikolic' - Arsovic'. FIDE has since rescinded that modification to the rule. The longest decisive tournament game is Fressinet-Kosteniuk, Villandry 2007, which Kosteniuk won in 237 moves.[2] The last 116 moves were a rook and bishop versus rook ending, as in Nikolic' - Arsovic' -- Why does he stagger his mind with the mathematics of the sky? Once the question mark has arisen in the human brain the answer must be found, if it takes a hundred years. A thousand years. -- Walter Reisch === Subject: Re: Every Chess Configuration posting-account=DLD3MQkAAACJxulKt9xouw3DPpXKssCI 1.1.4322; .NET CLR 1.0.3705; .NET CLR 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) On Jul 1, 9:33æam, rober...@ibd.nrc-cnrc.gc.ca (Walter Roberson) >and between these capture/recaptures, there could be many moves >without a capture, in the end game at least. So 150-move >end-games seem plausible, unless there's a forced mate >by one side in less than 150 moves. Wikipedia indicates, æ The longest tournament chess game ever to be played under modern time > æ rules was Nikolic'-Arsovic', Belgrade 1989, which lasted for 20 hours > æ and 15 minutes, ending in a 269-move draw.[1][2] At the time this game > æ was played, FIDE had modified the fifty move rule to allow 100 moves to > æ be played without a piece being captured in a rook and bishop versus > æ rook endgame, the situation in Nikolic' - Arsovic'. FIDE has since > æ rescinded that modification to the rule. æ The longest decisive tournament game is Fressinet-Kosteniuk, > æ Villandry 2007, which Kosteniuk won in 237 moves.[2] The last 116 > æ moves were a rook and bishop versus rook ending, as in Nikolic' - > æ Arsovic' In order to enumerate *all* chess games, you also have to list all the stupid ones. P.S., here is a 300+ move chess game from a contest {the actual game was longer, but the PGN tools cannot handle games longer than 300 full moves}: [Event CCRL 40/40] [Site CCRL] [Round 1.46.572] [White Chess Tiger 2007] [Black Rybka 1.0 64-bit] [Result 0-1] [WhiteElo 2792] [BlackElo 2921] [ECO E12] [Opening Queen's Indian defence] [PlyCount 600] 1.d4 Nf6 2.c4 e6 3.Nf3 b6 4.Nbd2 Bb4 5.a3 Bxd2+ 6.Qxd2 O-O 7.e3 Ne4 8.Qc2 Bb7 9.Bd3 f5 10.O-O Nc6 11.b4 Ne7 12.Nd2 Nf6 13.f3 Qe8 14.Nb3 Qh5 15.a4 d6 16.a5 e5 17.a6 Bc6 18.Be2 Bd7 19.b5 Ng6 20.f4 Qh6 21.Rf3 exf4 22.exf4 Qh4 23.g3 Qh3 24.d5 Rae8 25.Nd4 Re7 26.Bf1 Qh5 27.Qg2 Rfe8 28.Be2 Qh6 29.h3 Re4 30.Bb2 R4e7 31.Bd3 Qh5 32.Qh2 Nh8 33.Raf1 Ng6 34.R3f2 Re3 35.Be2 Qh6 36.Rf3 R3e7 37.Bd1 Nh8 38.Bb3 Nf7 39.Bc2 Qh5 40.Qg2 Nh6 41.Bd1 Qg6 42.Rc3 Kh8 43.Rb3 Qf7 44.Bc2 Kg8 45.Rc3 Kh8 46.Rcf3 Qh5 47.Bd1 Qf7 48.R3f2 Qg6 49.Re2 Rxe2 50.Bxe2 Ne4 51.Kh2 Ng8 52.Bd3 Ngf6 53.Re1 Kg8 54.Bc2 Qh6 55. Re2 Ng4+ 56.Kg1 Ngf6 57.Re3 Kf8 58.Bb1 Kg8 59.Bd3 Kf8 60.Bc2 Kg8 61.Nf3 Nh5 62.Bxe4 Rxe4 63.Bd4 Qg6 64.Kh2 Rxe3 65.Bxe3 Qe8 66.Bg1 h6 67.Qc2 Nf6 68.Bd4 Qh5 69.Qg2 Ne4 70.Bb2 Kf8 71.Qf1 g5 72.Bc1 g4 73.Nd4 Qg6 74.h4 Qe8 75.Qe2 Nc5 76.Qxe8+ Kxe8 77.Bd2 Ne4 78.Be1 Nc5 79.Nc2 Nd3 80.Bc3 Nc5 81. Kg2 Ke7 82.Kf1 Kf7 83.Ke2 Ne4 84.Be1 Ke7 85.Nd4 h5 86.Kd3 Nc5+ 87.Ke3 Ne4 88.Ne2 Kf7 89.Bc3 Kf8 90.Bd4 Kf7 91.Bb2 Ke8 92.Bg7 Ke7 93.Ba1 Ke8 94.Bb2 Ke7 95.Bg7 Ke8 96.Bh6 Kd8 97.Bg5+ Kc8 98.Kd3 Be8 99.Bh6 Kd7 100.Bg7 Ke7 101.Kd4 Kf7 102.Bh6 Ke7 103.Ke3 Kd8 104.Bf8 Bd7 105.Kd4 Ke8 106.Bh6 Kf7 107.Bg5 Ke8 108.Kd3 Nc5+ 109.Kc2 Ne4 110.Kb2 Bc8 111.Bh6 Kf7 112.Kb3 Bd7 113.Bg5 Ke8 114.Kc2 Bc8 115.Bh6 Kf7 116.Kd3 Bd7 117.Bg5 Ke8 118.Kd4 Bc8 119.Bh6 Ke7 120.Kd3 Bd7 121.Kc2 Kf7 122.Kc1 Ke7 123.Bg7 c6 124.bxc6 Bc8 125.Kc2 Bxa6 126.Kd3 b5 127.Bc3 bxc4+ 128.Ke3 Bc8 129.Be1 Nc5 130.Kd2 Nd3 131.Kd1 Kd8 132.Ba5+ Ke7 133.Kc2 Nf2 134.Be1 Nd3 135.Bd2 Kd8 136.Nc3 Nf2 137.Kc1 Nd3+ 138.Kb1 Nf2 139.Kc2 Ke7 140.Kb2 Ne4 141.Be1 Nc5 142.Ka3 Ne4 143.Kb4 Kd8 144.Kb5 Nc5 145.Bf2 Ne4 146.Nxe4 fxe4 147.Kxc4 Bf5 148.Bxa7 Kc7 149.Kb5 Bg6 150.Bb6+ Kc8 151.Be3 Kc7 152.Kc4 Bf5 153.Bf2 Bh7 154.Kb5 Bg6 155.Bd4 Bf5 156.Bb6+ Kb8 157.Kc4 Bc8 158.Kc3 Bf5 159.Be3 Kc7 160.Bc1 Bg6 161.Kd4 Bf5 162.Ba3 Bh7 163.Bb4 Bf5 164.Ba5+ Kc8 165.Bd2 Kc7 166.Bb4 Bg6 167.Ba5+ Kc8 168.Bc3 Kd8 169.Bb2 Kc7 170.Ke3 Kb6 171.Ba3 Kc7 172.Bb4 Bh7 173.Ba5+ Kc8 174.Bc3 Kc7 175.Bb4 Bf5 176.Ba5+ Kc8 177.Bb6 Bg6 178.Kd4 Bh7 179.Kc4 Bg6 180.Kc3 Bf5 181.Ba7 Kc7 182.Be3 Bg6 183.Bf2 Bf5 184.Ba7 Bg6 185.Kb4 Bf5 186.Be3 Bg6 187.Bd4 Bf5 188.Bf2 Bg6 189.Bg1 Bf7 190.Kc4 Bg6 191.Kd4 Bf5 192.Ke3 Bg6 193.f5 Bxf5 194.Bf2 Bh7 195.Be1 Bf5 196.Kf4 Bg6 197.Bb4 e3 198.Kxe3 Bf7 199.Ba5+ Kc8 200.Kd4 Bg6 201.c7 Bc2 202.Ke3 Ba4 203.Kf4 Be8 204.Kf5 Bb5 205.Ke6 Ba6 206.Kxd6 Bc4 207.Ke6 Bb5 208.d6 Bd7+ 209.Ke7 Bc6 210.Bc3 Bd7 211.Kf7 Bc6 212.Kf8 Bd7 213.Bh8 Be6 214.Ke7 Bf5 215.Bf6 Bd7 216.Kf7 Bf5 217.Ke8 Bd7+ 218.Kf8 Bb5 219.Bd4 Kd7 220.Kg7 Kc8 221.Ba1 Be8 222.Kf6 Bb5 223.Bc3 Bd7 224.Bd2 Bc6 225.Bh6 Bd7 226.Be3 Bc6 227.Kg5 Be8 228.Bd2 Kd7 229.Kh6 Bf7 230.Kg7 Be8 231.Bc1 Kc8 232.Bf4 Bd7 233.Kg6 Be8+ 234.Kf6 Bd7 235.Ke7 Ba4 236.Be3 Bb5 237.Kf7 Bd7 238.Bg1 Bb5 239.Bc5 Bd7 240.Kf8 Bb5 241.Be3 Kd7 242.Kf7 Kc8 243.Bh6 Bd7 244.Bg5 Bc6 245.Bf4 Bd7 246.Be5 Ba4 247.Kg6 Be8+ 248.Kh6 Kd7 249.Kg7 Kc8 250.Bb2 Bd7 251.Kf7 Bc6 252.Kg6 Be8+ 253.Kf6 Bb5 254.Kg5 Be8 255.Bg7 Kd7 256.c8=Q+ Kxc8 257.Be5 Kd8 258.Kf5 Kd7 259.Kf6 Kd8 260.Bc3 Kd7 261.Bb4 Kc6 262.Ke6 Bd7+ 263.Ke7 Bf5 264.Bd2 Bc8 265.Be1 Bd7 266.Bc3 Bc8 267.Bh8 Bf5 268.Bd4 Bd7 269.Be5 Bf5 270.Bg7 Bc8 271.Ba1 Bf5 272.Bf6 Bc8 273.Bb2 Bf5 274.Be5 Bd7 275.Bd4 Bf5 276.Bb2 Bd7 277.Bc1 Bf5 278.Bh6 Bc8 279.Be3 Bd7 280.Bb6 Bf5 281.Kf6 Bd7 282.Bc7 Be8 283.Ke7 Bd7 284.Ba5 Bf5 285.Bb4 Bd7 286.Ba3 Bf5 287.Bc1 Bc8 288.Bd2 Bd7 289.Bf4 Bf5 290.Kf7 Bd7 291.Kf6 Be8 292.Ke6 Bd7+ 293.Ke5 Be8 294.Bh6 Kd7 295.Bf8 Bg6 296.Kf6 Be8 297.Be7 Kc6 298.Ke6 Bd7+ 299.Ke5 Kc5 300.Bd8 Kc6 0-1 === Subject: The Definition of Proper Acceleration posting-account=lBRURwoAAAB_-Q_b04pGziaymfr5yRFx Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > This force is given by the equation g(x')=-(c^2)/x'. > I am afraid this formula doesn't make any sense to me. If the rocket > observer defines x' as distance from him, the formula tells him that > he must always experience infinite acceleration. > The x' coordinate is defined to be the Lorentz invariant quantity > (x')^2 =x^2-(ct)^2 Well that formula doesn't make any sense either, since it tells us > that x' is independent of g. The acceleration g is a function of x' so it doesn't have to appear explicitly in the definition of x'. Here is a straightforward method for finding that function. Let x' = D represent a fixed point in the accelerating coordinate system. Suppose D > 0. Then x^2 = D^2 + (ct)^2 Please note that x=x(t). Clearly, x(t) = D sqrt(1+ (ct)^2/D^2) represents the motion of the point x' = D in terms of time t. Note that x(0) = D. Let b = c^2/D^2 Then x(t) = D sqrt(1+ bt^2). Using baby calculus, we may expand sqrt(1+ bt^2) using the Maclaurin series. Thus: sqrt(1+ bt^2) = 1 + b(t^2)/2 - (b^2)(t^4)/8 + ... So x(t) = D + (c^2/2D)t^2 - (c^4/8D^3)t^4 + ... If time t is sufficiently small compared to c^2/2D, we may ignore the fourth and all higher orders of t in the expansion. Now recall your high school physics. The equation x = (1/2)at^2 represents distance traveled in time t when an object undergoes a constant acceleration a. Therefore a = c^2/D is a sensible Newtonian approximation for the initial acceleration of the point x' = D. Now consider what it means for the equation D^2 = x^2-(ct)^2 to be invariant under the Lorentz transformation. It means that whatever speed the point D attains in time t in one frame of reference, we can simply go to the co-moving frame where acceleration begins afresh from zero velocity at a new time t'=0 and compute the acceleration from there by merely repeating the calculation already performed. Therefore the acceleration a(x') = c^2/x' is well-defined. Shubee http://www.everythingimportant.org/relativity/special.pdf === Subject: Porn Addiction posting-account=XypZRAoAAABcJ6jzdk_DECEdagSV0nR_ Help, I'm addicted to porn. It's been going on for a while, but now I find myself surfing porn all day, downloading tons of material, and masturbating to it. I've even maxed out my credit card paying for sex sites and I've missed whole days from work because of this addiction. This is ruining my life and my health. What can I do to stop this?? TIA for any suggestions. === Subject: Re: Porn Addiction posting-account=Ac0S6wkAAAD2EX3Sovet7dRort-OEJoX 5.1),gzip(gfe),gzip(gfe) I too have a porn addiction and, in between all the orgasms, it used to worry me greatly. But then, I discovered a television program called Big Brother. I still have the porn addiction but now it doesn't worry me at all as clearly, if there are people that watch Big Brother, it means there really are people out there with even less going for them in their lives then I have. I find this to be of great comfort. === Subject: Re: Porn Addiction posting-account=tCEoyAoAAAAkltU5zxOoI8uJ4lyz5-kv .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; .NET CLR 3.5.21022),gzip(gfe),gzip(gfe) > I too have a porn addiction and, in between all the orgasms, it used > to worry me greatly. > Have a conversation with a stripper. That should cure you of your fascination with women, permanently. The problem of course, is that after having had an actual conversation with a stripper, there is the very real possibility that you might turn gay. === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? TIA for any suggestions. Take up math. === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? > TIA for any suggestions. Take up math. But he might then turn to mathturbation! -- Moving things in still pictures! === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? TIA for any suggestions. > Take up math. But he might then turn to mathturbation! Does that work somehow differently - or is it just the same thing said by someone with a lisp? === Subject: Re: Porn Addiction > Take up math. > But he might then turn to mathturbation! Does that work somehow differently - or is it just the same thing said by >someone with a lisp? Differently. You have to work it out with a pencil. -- Cynic === Subject: Re: Porn Addiction > Take up math. > But he might then turn to mathturbation! > Does that work somehow differently - or is it just the same thing said by > someone with a lisp? Differently. You have to work it out with a pencil. Gentlemen, that reminds me ... If a mathematician suffers from constipation, does he work it out with a slide-rule? > === Subject: Re: Porn Addiction Originator: fburton@nyx.net (Francis Burton) > Differently. You have to work it out with a pencil. Gentlemen, that reminds me ... If a mathematician suffers from constipation, does he work it out with a >slide-rule? Not logs? ;-) Francis === Subject: Re: Porn Addiction <486a86aa$0$2913$fa0fcedb@news.zen.co.uk> <1215007256.731668@irys.nyx.net> posting-account=XU2EOQoAAAAiZZKz1lHe5HzSjaGnJm77 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) > Differently. æYou have to work it out with a pencil. >Gentlemen, that reminds me ... >If a mathematician suffers from constipation, does he work it out with a >slide-rule? Not logs? ;-) Francis No, he just visits an enema porn site and uses his credit card to arrange a home visit from a sexy nurse. === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? > TIA for any suggestions. > Take up math. > But he might then turn to mathturbation! Does that work somehow differently - or is it just the same thing said by > someone with a lisp? Well, you can make use of your fingers for both! -- Moving things in still pictures! === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? > TIA for any suggestions. Take up math. I don't see OP's post, so I reply here. I gather this addiction can break up marriages and cause other problems. To OP, I might suggest taking a look at this Web Board: http://www.no-porn.com/board.html === Subject: Re: Porn Addiction > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? > TIA for any suggestions. Take up math. A great way to completely lose one's sex drive. Unless you come across a problem that's NP hard, of course. === Subject: Re: Porn Addiction <1wdak.77276$7m7.54907@newsfe30.ams2> posting-account=XU2EOQoAAAAiZZKz1lHe5HzSjaGnJm77 .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506),gzip(gfe),gzip(gfe) Get a girlfriend and stick your nose up her arse. === Subject: Re: Porn Addiction posting-account=c-RtwAoAAABkmy_eTS7qXcTbN5xGGVps Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe) > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? TIA for any suggestions. http://www.addictions.co.uk/addiction.asp?id=sex WM www.critest.com === Subject: Re: Porn Addiction posting-account=2dDofAoAAACla-mFMj57kvJ-zRtgGdse Gecko/20070914 Firefox/2.0.0.7,gzip(gfe),gzip(gfe) > Help, I'm addicted to porn. It's been going on for a while, but now I > find myself surfing porn all day, downloading tons of material, and > masturbating to it. I've even maxed out my credit card paying for sex > sites and I've missed whole days from work because of this addiction. > This is ruining my life and my health. What can I do to stop this?? TIA for any suggestions. INSTALL A PORN FILTER ON YOUR COMPUTER. This is the way you block explicit content from your computer. You should have it on every PC you use regularly. Install it from here: http://www.optenetpc.com/porn-filter.html Change the password to something you won't easily remember, so you won't be tempted to circumvent the restrictions. Excessive pornography use affects a significant number of internet users. However there is hope and you can overcome this. Good luck and let us know how you make out. === Subject: Mysterious calculation error 17^13 + 1 - 17^13 = 0 I have a pocket calculator that gives for 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. But it is somehow also very mysterious because IMO 17^13 is, relatively seen, not that a big number to hit some internal limits. I mean IMO it cannot be an overflow error. Another mystery: doing the same calculation on a PC in a programming language (MS C++) gives exactly the same buggy result!!! This is unbelievable! So then why this error, and what kind of math error is this? === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. æThis is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? If your calculator gave a result of x.xxxxxxExx then it has rounded off the low digits as it does not have enough room for them, so it thinks 17^13 + 1 is the same as 17^13, hence 0 as the result. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 posting-account=b4rQkQoAAADqqftYxNCieauSttMs6_zU 1.1.4322; InfoPath.1),gzip(gfe),gzip(gfe) > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. æThis is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? I found another calculator bug. When I ask my calculator to compute the square root of 2 it yields 1.41421356237 - which is a *rational* number! Unbelievable! Even the ancient Greeks knew that the square root of 2 is irrational. Maybe I should stop using this piece of junk and pull out my old slide rule instead. The digital is just too deceptive. -scattered p.s. A less facetious but probably well-known example: (1- cos(0.0000001))/0.0000001^2 evaluates to 0 on most (all?) calculators but a simple L'Hospital's rule calculation shows that the answer should be very close to 0.5 === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 Adem24 a .8ecrit : > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. is IMO some sort of certification I have not heard of ? Ok, sarcasm aside, it *is*, of course, an kind of overflow error : your calculator simply cannot take care of enough significant digits to see the difference between A and A+1, as soon as A is greater than 10^12 or so... > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? > === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 Adem24 a ?crit : > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. Unless your calculator works in exact integer arithmetic up to at least 16 significant digits, as mind does, you will get round off error when adding or subtracting numbers of such different sizes. On an hp50 in exact mode, 17^13 + 1 - 17^13 evaluates to 1. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 posting-account=yxbZkgkAAABQBvyYeebYQ-PAvi0uT3tG Gecko/20071127 Firefox/2.0.0.11,gzip(gfe),gzip(gfe) Adem24 acrit : > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. Unless your calculator works in exact integer arithmetic up to at least > 16 significant digits, as mind does, you will get round off error when > adding or subtracting numbers of such different sizes. On an hp50 in exact mode, 17^13 + 1 - 17^13 evaluates to 1. The MS/Windows Calculator also gives an exact result: 17^13 = 9904578032905937 It appears to handle mantissas of around 106 bits or so. 2^106 = 81129638414606681695789005144064 === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 - > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? > Try (17.0)^13 + 1.0 - (17.0)^13; then the evaluation will be performed with reals instead of integers. This could make all the difference. 17^13 = = (1 + 16)^13 = = 1 + (13 choose 1).16 + (13 choose 2).16^2 + ... + (13 choose 12).16^12 + 16^13 = = about 9.905 * 10^15 It occupies 54 bits. It might be the case that 17^13 with 17 and 13 entered without decimal points is treated in the hardware as an integer. Assume binary hardware. Try and evaluate this with integers occupying no more than 16 bits, 32 bits, 48 bits. When working modulo 2^16 = 16^4 = 65536 only four terms do not vanish. Modulo 2^32 = 16^8 only eight terms do not vanish. All this dose not yet explain how the term 1 can get lost in the calculation process. There are at least two possibilities: (A) The hardware switches somewhere in the process automatically and without warning to floating-point. Then unity can very well be non-significant with respect the number 17^13 already present in the floating-point register. (B) The hardware is buggy. Good luck: Johan E. Mebius === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 - posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. æThis is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? Try (17.0)^13 + 1.0 - (17.0)^13; then the evaluation will be performed with reals instead > of integers. This could make all the difference. 17^13 = > = (1 + 16)^13 = > = 1 + (13 choose 1).16 + (13 choose 2).16^2 + ... + (13 choose 12).16^12 + 16^13 = > = about 9.905 * 10^15 > It occupies 54 bits. > Actually you don't even need that. You just need log2(17^13) = 13 log2(17) ~ 13 * 4.1 ~ 53.3, which is greater than 53. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 - Nntp-Posting-Host: hera.cwi.nl > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? > > Try (17.0)^13 + 1.0 - (17.0)^13; then the evaluation will be performed > with reals instead of integers. This could make all the difference. Will not be the case. Most pocket calculators use (internally) decimal floating point with a specific number of digits in the mantissa. If the number of digits is not larger than 15 the 1 will be rounded away in the addition. > (A) The hardware switches somewhere in the process automatically and > without warning to floating-point. Then unity can very well be > non-significant with respect the number 17^13 already present in > the floating-point register. That is approximately the case with most pocket calculators. If you have one that does not show a zero result for 1/3, it almost certainly uses floating-point internally. With that result you can find the number of digits it uses internally: (1) set counter to 0 (2) add 1 to counter (3) multiply by 10 and subtract 3 (4) if the result is non-zero goto step 2 The number you find this way (counter) is currently on most calculators larger than the number actually displayed. If the calculator uses binary operations rather than decimal at some stage your display will show decimal digits differing from 3, and the process possibly does not stop. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. In that case, your opinion is worthless. Please get some better opinions. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! Au contraire. That result is far from unexpected, and entirely in conformance with the C++ standard. > This is unbelievable! Your belief system is flawed. Get a better one. > So then why this error, and what kind of math error is this? Your inability to understand the internals of the machine and programs you use. Phil -- -- Microsoft voice recognition live demonstration === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 posting-account=fwSgtAkAAACFnX70ssKwbvm9_oCZVHrx 2.0.50727; .NET CLR 3.0.04506.648; .NET CLR 3.5.21022; .NET CLR 1.1.4322),gzip(gfe),gzip(gfe) On Jun 30, 3:47æpm, Phil Carmody I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. æThis is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. In that case, your opinion is worthless. Please get > some better opinions. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! Au contraire. That result is far from unexpected, and > entirely in conformance with the C++ standard. > This is unbelievable! Your belief system is flawed. Get a better one. > So then why this error, and what kind of math error is this? Your inability to understand the internals of the > machine and programs you use. > Most likely he is able to understand it, he just hasn't been given that understanding. Although if he read these posts, I'm sure he's gotten it by now. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. 17^13 ~ 10^16. Most pocket calculators cannot handle 16 decimal digits precisely. It is, in fact, an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly I'm guessing you are using the C++ type |double| which gives you a 52 bit mantissa. 2^53 < 17^13 < 2^54, so that means that the 1 is lost in precision. If you use |long double|, I believe you should have the necessary precision. > This is unbelievable! > So then why this error, and what kind of math error is this? You've invoked a cardinal rule of numerical analysis: Do not subtract two numbers whose value is near equal. You lose precision in gobs; in this case, the entire precision the computer/calculator knows of the number. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow > error. Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > This is unbelievable! > So then why this error, and what kind of math error is this? 17^13 = 9904578032905937 That's 16 digits. How many digits does your calculator maintain? Probably only 12 (10 displayed plus two more kept internally). You probably used single precision arithmatic in your C++ program. Try again using double precision. === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 posting-account=OKTeIQkAAAAZk6JK1hK7-grwpoUDNy98 CLR 1.1.4322; .NET CLR 2.0.50727; InfoPath.1),gzip(gfe),gzip(gfe) > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. æThis is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! Because you used MS C++. Only a fool would do that. > This is unbelievable! Not for MS C++. > So then why this error, and what kind of math error is this? It's not. It's a programming language error. Use one not prone to such errors, like Python: IDLE 1.2 > a = 17**13 > a 9904578032905937L > a + 1 - a 1L === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 > > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! Because you used MS C++. Only a fool would do that. > This is unbelievable! Not for MS C++. But also the Casio fx-115WA pocket calculator gives the same buggy result! I wonder why? > So then why this error, and what kind of math error is this? It's not. It's a programming language error. > Use one not prone to such errors Sure, but that's not the question here. If one discovers such an error then one better analyses it first throughly before deciding what to do next... The mystery here is that the error is in two different, AFAIK independent products from different companies... === Subject: Re: Mysterious calculation error 17^13 + 1 - 17^13 = 0 > I have a pocket calculator that gives for > 17^13 + 1 - 17^13 = 0. This is obviously a buggy result. > But it is somehow also very mysterious because > IMO 17^13 is, relatively seen, not that a big number > to hit some internal limits. I mean IMO it cannot be an overflow error. > Another mystery: doing the same calculation on a PC > in a programming language (MS C++) gives exactly > the same buggy result!!! > Because you used MS C++. Only a fool would do that. > This is unbelievable! > Not for MS C++. But also the Casio fx-115WA pocket calculator gives the same buggy result! >I wonder why? > So then why this error, and what kind of math error is this? > It's not. It's a programming language error. > Use one not prone to such errors Sure, but that's not the question here. >If one discovers such an error then one better analyses it first >throughly before deciding what to do next... >The mystery here is that the error is in two different, >AFAIK independent products from different companies... 53 bits or less (such as the IEEE double-precision floating point format's mantissa field, for example) is probably used in both cases. --Spud Demon === Subject: How low can you jackasses get. I have a physics PhD. Get over it Cc: rwill9955b@yahoo.com posting-account=OprVJAoAAAAf9u6DK4OBqwBaLD1BPEld 5.1),gzip(gfe),gzip(gfe) Once again, this worthless group went to a new low. I was alledged to have falsely claimed to have a physics PhD FYI, I never championed James Harris. I tried to help him and quit FYI, if you are so stupid as to claim I do not have a physics PhD, well them, try buying Brooklyn bridge. Its easier than trying to change a known academic credential. Want to prove I do NOT have a physics PhD ? If you have the gonads to try, after I nuke you for it, I hope you realize that some things are not changable on YOUR part, bigotry on your part non withstanding. Have fun, you KKK style bigots. Andrew Beckwith === Subject: Wikimization posting-account=TCOH9AoAAADiHysQ5xKKV-UN-ApUH0nV 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) There's a resource on the web for the purpose of collaboration when writing mathematical papers. It is called http://Wikimization.org Essentially, the idea is this: If two or more authors wish to collaborate, they must set up some kind of personal semaphore system to make sure all authors are not editing simultaneously. This can be quite inefficient. Suppose there were a place on the web where all the authors could go to edit their joint work. That place should have a history buffer to show who was doing what and when. It should also have a LaTeX-style input so the work can be easily compiled later when ready for publication. Wikimization is a meld of the two words Wiki (the Hawaiian word for quick) and Optimization. The site is not only meant to be a repository of finished work - rather, it is intended as a mecca for work in progress, available to each author anywhere there is Internet access. Pages can be protected so that your work is not hacked. Like a Wiki, comparison of any two previous revisions is instantaneous via the History buffer, and reversion to a previous version is always possible. I think Wikimization's most powerful attribute is the ability to write math in LaTeX, a capability absent from most all newsgroups. Try it out and let me know what you think. Jon === Subject: Re: Wikimization > There's a resource on the web for the purpose of collaboration when > writing mathematical papers. > It is called http://Wikimization.org Essentially, the idea is this: If two or more authors wish to > collaborate, they must set up some kind of personal semaphore system > to make sure all authors are not editing simultaneously. This can be > quite inefficient. Suppose there were a place on the web where all > the authors could go to edit their joint work. That place should > have > a history buffer to show who was doing what and when. It should also > have a LaTeX-style input so the work can be easily compiled later > when > ready for publication. > Source control systems used by software developers solved this problem long ago. There are two basic approaches: * check-in -- anyone can edit to their hearts content. At check-in time, if conflicts are found with the current revision, the software makes a (usually pretty decent) attempt to resolve the conflicts; otherwise humans have to resolve the conflicts by hand. Subversion and CVS are two prominent check-in systems. * check-out -- only one person at a time can have a given file checked out with write access. Perforce is a prominent example of a check-out based system. === Subject: Re: Wikimization posting-account=TCOH9AoAAADiHysQ5xKKV-UN-ApUH0nV 2.0.50727; .NET CLR 3.0.04506.30; .NET CLR 3.0.04506.648),gzip(gfe),gzip(gfe) > There's a resource on the web for the purpose of collaboration when > writing mathematical papers. > It is called æhttp://Wikimization.org > Essentially, the idea is this: æIf two or more authors wish to > collaborate, they must set up some kind of personal semaphore system > to make sure all authors are not editing simultaneously. æThis can be > quite inefficient. æSuppose there were a place on the web where all > the authors could go to edit their joint work. æThat place should > have > a history buffer to show who was doing what and when. æIt should also > have a LaTeX-style input so the work can be easily compiled later > when > ready for publication. Source control systems used by software developers solved this problem > long ago. There are two basic approaches: * check-in -- anyone can edit to their hearts content. At check-in > time, if conflicts are found with the current revision, the software > makes a (usually pretty decent) attempt to resolve the conflicts; > otherwise humans have to resolve the conflicts by hand. Subversion and > CVS are two prominent check-in systems. * check-out -- only one person at a time can have a given file checked > out with write access. Perforce is a prominent example of a check-out > based system.- Hide quoted text - - Show quoted text - But with the source control method you cite, one cannot write mathematics in LaTeX and one cannot write HTML simultaneuously into the same document. Using Wikimization instead, the other authors don't have to wait for you to get around to releasing the document you checked out last week. They can watch to see the changes you are making in nearly real time. Source code (specifically for programmer-authors) is entered into Wikimization using HTML 
/pre> tags.
Jon
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Subject: Deconstructing affine transformations into rotation, scale, skew
Given a 2-dimensional affine transform (6 coefficients as defined at 
http://java.sun.com/j2se/1.5.0/docs/api/java/awt/geom/AffineTransform.html 
)
        [ x']   [  m00  m01  m02  ] [ x ]   [ m00x + m01y + m02 ]
        [ y'] = [  m10  m11  m12  ] [ y ] = [ m10x + m11y + m12 ]
        [ 1 ]   [   0    0    1   ] [ 1 ]   [         1         ]
How can I separate this transformation into its components?  
The translation vector is obviously , but after that it's not so
trivial.  I would like to come up with:
* Rotation (scalar)
* Scaling or dilation (scalar or 2-D vector)
* Indication of whether the transform flips the image (bit)
* Skew (2-D vector?)
Pointers to info or even just the right terms to use in my web search
will be greatly appreciated!
--Spud Demon
===
Subject: Re: Deconstructing affine transformations into rotation, scale, 
skew
        posting-account=Xop4ywkAAABVt57ERw7fduXnH3Hf8NRj
        Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe)
> Given a 2-dimensional affine transform (6 coefficients as defined 
athttp://java.sun.com/j2se/1.5.0/docs/api/java/awt/geom/AffineTransform...)
         [ x']   [  m00  m01  m02  ] [ x ]   [ m00x + m01y + m02 ]
>         [ y'] = [  m10  m11  m12  ] [ y ] = [ m10x + m11y + m12 ]
>         [ 1 ]   [   0    0    1   ] [ 1 ]   [         1         ]
 How can I separate this transformation into its components?
 The translation vector is obviously , but after that it's not so
> trivial.  I would like to come up with:
 * Rotation (scalar)
> * Scaling or dilation (scalar or 2-D vector)
> * Indication of whether the transform flips the image (bit)
> * Skew (2-D vector?)
 Pointers to info or even just the right terms to use in my web search
> will be greatly appreciated!
 --Spud Demon
Maybe this can help:
===
Subject: Re: Deconstructing affine transformations into rotation, scale, 
skew
        posting-account=xM691AkAAACRhg2rzU0Kd6hZjdls4krD
        rv:1.8.1.14) Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe)
> Given a 2-dimensional affine transform (6 coefficients as defined 
athttp://java.sun.com/j2se/1.5.0/docs/api/java/awt/geom/AffineTransform...)
         [ x']   [  m00  m01  m02  ] [ x ]   [ m00x + m01y + m02 ]
>         [ y'] = [  m10  m11  m12  ] [ y ] = [ m10x + m11y + m12 ]
>         [ 1 ]   [   0    0    1   ] [ 1 ]   [         1         ]
 How can I separate this transformation into its components?
 The translation vector is obviously , but after that it's not so
> trivial.  I would like to come up with:
 * Rotation (scalar)
> * Scaling or dilation (scalar or 2-D vector)
> * Indication of whether the transform flips the image (bit)
> * Skew (2-D vector?)
 Pointers to info or even just the right terms to use in my web search
> will be greatly appreciated!
 --Spud Demon
Look up singular value decomposition (SVD). You want the SVD of
m = [ m00  m01 ] = U D V'.
    [ m10  m11 ]
U and V are orthonormal, and represent rotations with possible
reflections. D is positive diagonal, and represents rescaling.
===
Subject: Physics: Principles with Applications, 6th Ed. by Giancoli
        posting-account=DD1ZUwoAAABWXmcCJq8SIX-ttQ6iunfQ
        3.2.0; FunWebProducts; .NET CLR 1.1.4322; .NET CLR 
2.0.50727),gzip(gfe),gzip(gfe)
Please i,m looking for a complete study guide for Physics: Principles
with Applications, 6th Ed. by Giancoli, Not the selected solution.
Please if you have it  contact me ASAP because i need it for the
summer 2008.
2BIG
===
Subject: A simple puzzle in number theory or combinatorics
Let's denote the binomial coefficient
x! / [ (x-y)! y! ]
by C(x,y). Let p be a prime number, n an integer >0, and m an integer >= 0. 

Prove that
C( p^n, mp ) - C( p^(n-1), m )
is divisible by p^n.
===
Subject: Re: A simple puzzle in number theory or combinatorics
Larry Hammick
> Let's denote the binomial coefficient
> x! / [ (x-y)! y! ]
> by C(x,y). Let p be a prime number, n an integer >0, and m an integer >= 
> 0. Prove that
> C( p^n, mp ) - C( p^(n-1), m )
> is divisible by p^n.
A superficially stronger statement:
Let n be an integer >=1 and denote the Mobius function by M. Define a 
polynomial
f(X) = sum_{d | n} M(d) (1 + X^d)^{n/d}
All the coefficients of f are divisible by n.
Hint: Let S be a cyclic group of n elements. Of the 2^n subsets of S, how 
many are not fixed by any non-zero translation?
I don't yet have a proof of this even stronger version:
Let g(X) be any polynomial in X over Z, fix n>=1, and define
f(X) = sum_{d | n} M(d) (g( X^d))^{n/d}
All the coefficients of f are divisible by n. 
===
Subject: Interchange of trace and integration

can the order of matrix trace operator and integration be
interchanged ? specifically, i've come across the following result in
a paper i'm reading :
int_X (x - mu_x)^T E^{-1} (x - mu_x) dP(x) = trace { C_x E^{-1} }
(in riemann-steljes form)
where,
C_x = E{(x - mu_x) (x - mu_x)^T}   ... the expectation operator
working thru this result, i've come to:
LHS = int_X p(x) trace{ (x - mu_x) (x - mu_x)^T E^{-1} } dP(x)
how do i proceed from here ?
-fj
ps: btw - i'm not confident of the stated result itself - the
terminology and notation in the paper is extremely obfuscated. does
the result make sense ?
===
Subject: Re: Interchange of trace and integration
 can the order of matrix trace operator and integration be
> interchanged ? specifically, i've come across the following result in
> a paper i'm reading :
I will assume you meant to write Can, Specifically, I've and
I'm, etc., etc. Similarly below.
 int_X (x - mu_x)^T E^{-1} (x - mu_x) dP(x) = trace { C_x E^{-1} }
 (in riemann-steljes form)
[Riemann-Stieltjes]
 where,
> C_x = E{(x - mu_x) (x - mu_x)^T}   ... the expectation operator
I assume /this/ E has nothing to do with the E^(-1) in the integrand.
Is that correct?
 working thru this result, i've come to:
 LHS = int_X p(x) trace{ (x - mu_x) (x - mu_x)^T E^{-1} } dP(x)
If the vectors finite dimensional, the result is easy. Say x and mu_x
are in R^n; let m = mu_x for notational simplicity. The integrand is
sum{i} sum{j} (x_i - m_i)G_{i,j} (x_j - m_j), where G = E^(-1).
Integrating, we get sum_{i}sum_{j} C_{j,i}*G_{i,j}, where C_{j,i} =
C_{i,j} = E(x_i-mu_i)(X_j - m_j). This is the result you want.
R.G. Vickson
 how do i proceed from here ?
 -fj
 ps: btw - i'm not confident of the stated result itself - the
> terminology and notation in the paper is extremely obfuscated. does
> the result make sense ?
===
Subject: Integer as the sum of four squares
        posting-account=HaopWgoAAADs72-s8RQYwP_-ruRUuNzX
        .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 3.0.04506; 
        InfoPath.1),gzip(gfe),gzip(gfe)
I am reading a book and it makes the following statement.
Statement: Every integer can be written as the sum of four squares.
This was conjectered by Fermat and proven by Lagrange.
The book I am reading does not have this proof.
Is it a complicated proof?
I don't have a copy of Herstein handy - am guessing maybe it is in
there.
Is there a proof online or can it be summarized/mapped out by someone?
~A
===
Subject: Re: Integer as the sum of four squares
        posting-account=a6woBRAAAADpNFZJBA7ZBx35zXaKmaP4
        Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe)
 I am reading a book and it makes the following statement.
 Statement: Every integer can be written as the sum of four squares.
 This was conjectered by Fermat and proven by Lagrange.
 The book I am reading does not have this proof.
 Is it a complicated proof?
 I don't have a copy of Herstein handy - am guessing maybe it is in
> there.
 Is there a proof online or can it be summarized/mapped out by someone?
> ~A
This seems to be a concise presentation of the standard
proof:
http://www.math.ualberta.ca/~mlalin/Lagrange.pdf
An identity of Euler's is used to reduce the proof to
the case of a prime p being the sum of four squares.
===
Subject: El Euro 2008 - Fiesta- Deportivos.com
        posting-account=QYyPBAoAAADQr3r4qq0j1FGc5ru4UlRX
        5.1),gzip(gfe),gzip(gfe)
Deberiamos organizar un grupo para celebrar el Euro 2008
En que pais estan la mayoria de fanaticos? dimelo aqui
Nos gustaria organizar una fiesta en el pais con mas miembros.
Manuel
Deportivos.com
===
Subject: Geometry without Calculus
        posting-account=NzomCgoAAADSxo8dItZFimQE_f4Fbqcn
        .NET CLR 2.0.50727; Media Center PC 5.0; .NET CLR 
3.0.04506),gzip(gfe),gzip(gfe)
Round geometry in any dimension is an unchanging curve.
Mitch Raemsch
===
Subject: Re: Geometry without Calculus
> Round geometry in any dimension is an unchanging curve.
 Mitch Raemsch
Have you forgotten to take your meds again?
Bob Kolker
===
Subject: JSH:  We have a problem
        posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif
        Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe)
Go local.  Message.  Finish it this time.  Clear.  Tag.
No permissions on the other.  Hanging on a limb.
Code.  Expedite.  Time it.
Finish.
___JSH
===
Subject: Re: JSH:  We have a problem
> Go local.  Message.  Finish it this time.  Clear.
>  Tag.
 No permissions on the other.  Hanging on a limb.
 Code.  Expedite.  Time it.
 Finish.
 
> ___JSH
Cool! It's some top-secret trigger phrases, no doubt for the benefit
of all those powerful underground contacts of James', who definitely
do exist and aren't just part of an infantile fantasy being lived
out in public.
Sure is a mystery how you ever ended up on that terrorist watchlist,
eh?
===
Subject: Re: We have a problem
> Go local.  Message.  Finish it this time.  Clear.  Tag.
 No permissions on the other.  Hanging on a limb.
 Code.  Expedite.  Time it.
 Finish.
> ___JSH
Monkey Math at its best!
http://www.stupidmonkeyplanet.com/wp-content/uploads/2007/08/monkey01.jpg
===
Subject: Re: JSH:  We have a problem
> Go local.  Message.  Finish it this time.  Clear.  Tag.
 No permissions on the other.  Hanging on a limb.
 Code.  Expedite.  Time it.
 Finish.
You. Confirmed. That. You Are. Crackpot. Goodbye.
 >/dev/null
===
Subject: Re: JSH: We have a problem
        posting-account=n1ZfDgkAAABbCs44qOtz8dP-RkWuEBif
        Gecko/2008052906 Firefox/3.0,gzip(gfe),gzip(gfe)
> Go local. æMessage. æFinish it this time. 
æClear. æTag.
 No permissions on the other. æHanging on a limb.
 Code. æExpedite. æTime it.
 Finish.
    JSH
Across.  Dog.  Triangulate.  Window.  Fingerprint.  Tag.
Finish.
===
Subject: Re: JSH: We have a problem
> Go local. Message. Finish it this time. Clear. Tag.
> No permissions on the other. Hanging on a limb.
> Code. Expedite. Time it.
> Finish.
> ___JSH
>Across.  Dog.  Triangulate.  Window.  Fingerprint.  Tag.
Finish.
TERMINATE.
EOF
END
STOP
QUIT
GOTO 10
to exault your ratings in life will ultimatly fail as the SW dude that 
hacked at that code was moved over to counting used vegatables)
===
Subject: rationalize
rationalize(denominator);
1/(3+Sqrt[7]+19^(1/3)+53^(1/4))
===
Subject: Re: rationalize
Q[X, Y, Z]/ < X^2 - 7, Y^3 - 19, Z^4 - 53 >
([3] + [X] + [Y] + [Z])*(_________) = [1]
(_________)=_________
===
Subject: Re: rationalize
ex;
http://www.teachertube.com/view_video.php?viewkey=617e013417f26f7d0f82
Q(i)
|
|
Q
===
Subject: Re: rationalize
<4859228.1214893767589.JavaMail.jakarta@nitrogen.mathforum.org>,
> rationalize(denominator);
> 1/(3+Sqrt[7]+19^(1/3)+53^(1/4))
Why? 
Will I learn something I don't already know? 
Will you learn something you don't already know? 
Is there anything that makes this question more interesting 
than, say, find 12345^67890?
-- 
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Test Bank & Solutions for Accounting & Finance Courses
        posting-account=EN0WoQoAAADp8f0Fz2nDBf5LRPdx1zjc
        CLR 1.1.4322),gzip(gfe),gzip(gfe)
hey please i need the answers for Principles of Marketing,
===
Subject: 1/x
f(x)=x^24 - 72*x^23 + 2400*x^22 - 49256*x^21 + 
  698562*x^20 - 7291608*x^19 + 
  58369116*x^18 - 369867600*x^17 + 
  1910072007*x^16 - 8311616312*x^15 + 
  32206729512*x^14 - 124442758416*x^13 + 
  551597255058*x^12 - 2739648536688*x^11 + 
  12734831511744*x^10 - 48158368209752*
   x^9 + 137282132781081*x^8 - 
  274413101966928*x^7 + 325898880067188*
   x^6 - 60063972905136*x^5 - 
  454476659236500*x^4 + 529620263872096*
   x^3 + 255710100957984*x^2 - 
  834834613593600*x + 230834314922192
---------------------------------------------
Q[x]/f(x)*Q[x]
1/x=__________________________________________
===
Subject: SWEDENBORG ON LONGITUDE
             SWEDENBORG ON LONGITUDE
   In 1721 (Baron) Emanuel Swedenborg published a simple
method for finding longitude from the moon.  It is known
that he personally discussed it in London with Royal
Astronomer Flamsteed and Edmond Halley.  Apparently neither
of them thought it erroneous or incompetent.
   Swedenborg of course later became world famous for
publishing visionary theological speculations.  According to
me he had a schizophrenic reaction at the age of 55 and gave
up routine science to conduct the world's first scientific
investigation of schizophrenic visions, and which resulted
in his celebrated book _Heaven and Hell_; the world's most
famous quasi-technical description of Life After Death.
   What I want to know, is how brilliant a scientist and
mathematician was he before he became fascinated with mental
phenomena?  To wit; how competent and workable was his
proposed method of finding Longitude by:
       observing the moon between two stars
as described at:
                     (two pages... click to second page)
   I have not obtained the original and this online version
has barely readable diagrams.
   However, it appears simple, and a knowledgeable
astrophysicist should be able to discern immediately what he
was doing and whether it would work.
   If anyone competent in this area could take 5 minutes to
look at this and recognize how it worked.... could you
explain it to this physicist in modern English and give an
opinion as to how astute Swedenborg actually was as a
scientist... bearing in mind he was only in his twenties
when he dreamed up this longitude method and believed that
it was a viable contender for the fabulous Longitude Prize
offered by the British government in 1714.
Very obliged, Hammond
     SCIENTIFIC PROOF OF GOD WEBSITE
 http://geocities.com/scientific_proof_of_god
   mirror site:
 http://proof-of-god.freewebsitehosting.com
      GOD=G_uv   (a folk song on mp3)
 http://interrobang.jwgh.org/songs/hammond.mp3
===
Subject: Re: SWEDENBORG ON LONGITUDE
        posting-account=RpyoBAoAAABcM4UEoRGnChTPDHd9tv5P
        Gecko/20080404 Firefox/2.0.0.14,gzip(gfe),gzip(gfe)
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?  To wit; how competent and workable was his
> proposed method of finding Longitude by:
>        observing the moon between two stars
> as described at:
                      (two pages... click to second page)
    I have not obtained the original and this online version
> has barely readable diagrams.
>    However, it appears simple, and a knowledgeable
> astrophysicist should be able to discern immediately what he
> was doing and whether it would work.
>    If anyone competent in this area could take 5 minutes to
> look at this and recognize how it worked.... could you
> explain it to this physicist in modern English and give an
> opinion as to how astute Swedenborg actually was as a
> scientist... bearing in mind he was only in his twenties
> when he dreamed up this longitude method and believed that
> it was a viable contender for the fabulous Longitude Prize
> offered by the British government in 1714.
> Very obliged, Hammond
Seems reasonable George.
However, the amount of machinery and difficulty in measuring the
precise angles necessary to determine the longitude are likely more
difficult than that of building a clock.
I've been meaning to read Longitude by Dava Sobel, probably some
better answers as to the politics and 1714 technology that went into
the Longitude prize.
===
Subject: Re: SWEDENBORG ON LONGITUDE
>[Hammond]
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?  To wit; how competent and workable was his
> proposed method of finding Longitude by:
>        observing the moon between two stars
> as described at:
>                      (two pages... click to second page)
>    I have not obtained the original and this online version
> has barely readable diagrams.
>    However, it appears simple, and a knowledgeable
> astrophysicist should be able to discern immediately what he
> was doing and whether it would work.
>    If anyone competent in this area could take 5 minutes to
> look at this and recognize how it worked.... could you
> explain it to this physicist in modern English and give an
> opinion as to how astute Swedenborg actually was as a
> scientist... bearing in mind he was only in his twenties
> when he dreamed up this longitude method and believed that
> it was a viable contender for the fabulous Longitude Prize
> offered by the British government in 1714.
> Very obliged, Hammond
>[Luke]
>Seems reasonable George.
However, the amount of machinery and difficulty in measuring the
>precise angles necessary to determine the longitude are likely more
>difficult than that of building a clock.
I've been meaning to read Longitude by Dava Sobel, probably some
>better answers as to the politics and 1714 technology that went into
>the Longitude prize.
[Hammond]
   I haven't studied it but all lunar distance methods are
based on using the moon's motion around the Zodiac  as a
clock.  Problem is the moon only moves around the Zodiac
once a month while the minute hand of a clock rotates once
every hour meaning the moon moves (30)(24)=720 times slower
than a clock.
   This would indicate that very precise measurements of the
moon would be necessary.  Apparently however, the genius of
Swedenborg's method is that it uses the alignment of the
moon between 2 stars whose positions are already known and
therefore simple alignment eliminatesw most of the accurate
measurements otherwise required.
   I wish I could get a knowlegable opinion of his scheme...
whch is probably already well known to astronomical experts
somwhere.
     SCIENTIFIC PROOF OF GOD WEBSITE
 http://geocities.com/scientific_proof_of_god
   mirror site:
 http://proof-of-god.freewebsitehosting.com
      GOD=G_uv   (a folk song on mp3)
 http://interrobang.jwgh.org/songs/hammond.mp3
===
Subject: Re: SWEDENBORG ON LONGITUDE
        <1kgn641q7oesceoeq3v96tbes2i8ikiida@4ax.com>
        posting-account=vyjRTQgAAAAlmipmdUnA2I7cIRBWlv7_
        Gecko/20061029 SeaMonkey/1.0.6,gzip(gfe),gzip(gfe)
>[Hammond]
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?  To wit; how competent and workable was his
> proposed method of finding Longitude by:
>        observing the moon between two stars
> as described at:
 >                      (two pages... click to second page)
 >    I have not obtained the original and this online version
> has barely readable diagrams.
>    However, it appears simple, and a knowledgeable
> astrophysicist should be able to discern immediately what he
> was doing and whether it would work.
>    If anyone competent in this area could take 5 minutes to
> look at this and recognize how it worked.... could you
> explain it to this physicist in modern English and give an
> opinion as to how astute Swedenborg actually was as a
> scientist... bearing in mind he was only in his twenties
> when he dreamed up this longitude method and believed that
> it was a viable contender for the fabulous Longitude Prize
> offered by the British government in 1714.
> Very obliged, Hammond
 >[Luke]
>Seems reasonable George.
 >However, the amount of machinery and difficulty in measuring the
>precise angles necessary to determine the longitude are likely more
>difficult than that of building a clock.
 >I've been meaning to read Longitude by Dava Sobel, probably some
>better answers as to the politics and 1714 technology that went into
>the Longitude prize.
> [Hammond]
>    I haven't studied it but all lunar distance methods are
> based on using the moon's motion around the Zodiac  as a
> clock.  Problem is the moon only moves around the Zodiac
> once a month while the minute hand of a clock rotates once
> every hour meaning the moon moves (30)(24)=720 times slower
> than a clock.
>    This would indicate that very precise measurements of the
> moon would be necessary.  Apparently however, the genius of
> Swedenborg's method is that it uses the alignment of the
> moon between 2 stars whose positions are already known and
> therefore simple alignment eliminatesw most of the accurate
> measurements otherwise required.
>    I wish I could get a knowlegable opinion of his scheme...
> whch is probably already well known to astronomical experts
> somwhere.
Hi
 The problem with this method was that the Moon was not always
seen at night. Ships can't wait for the Moon.
 Using a clock, one could use Moon, Sun or stars.
Dwight
===
Subject: Re: SWEDENBORG ON LONGITUDE
>[Hammond]
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?  To wit; how competent and workable was his
> proposed method of finding Longitude by:
>        observing the moon between two stars
> as described at:
>                      (two pages... click to second page)
>    I have not obtained the original and this online version
> has barely readable diagrams.
>    However, it appears simple, and a knowledgeable
> astrophysicist should be able to discern immediately what he
> was doing and whether it would work.
>    If anyone competent in this area could take 5 minutes to
> look at this and recognize how it worked.... could you
> explain it to this physicist in modern English and give an
> opinion as to how astute Swedenborg actually was as a
> scientist... bearing in mind he was only in his twenties
> when he dreamed up this longitude method and believed that
> it was a viable contender for the fabulous Longitude Prize
> offered by the British government in 1714.
> Very obliged, Hammond
>[Luke]
>Seems reasonable George.
>However, the amount of machinery and difficulty in measuring the
>precise angles necessary to determine the longitude are likely more
>difficult than that of building a clock.
>I've been meaning to read Longitude by Dava Sobel, probably some
>better answers as to the politics and 1714 technology that went into
>the Longitude prize.
> [Hammond]
>    I haven't studied it but all lunar distance methods are
> based on using the moon's motion around the Zodiac  as a
> clock.  Problem is the moon only moves around the Zodiac
> once a month while the minute hand of a clock rotates once
> every hour meaning the moon moves (30)(24)=720 times slower
> than a clock.
>    This would indicate that very precise measurements of the
> moon would be necessary.  Apparently however, the genius of
> Swedenborg's method is that it uses the alignment of the
> moon between 2 stars whose positions are already known and
> therefore simple alignment eliminatesw most of the accurate
> measurements otherwise required.
>    I wish I could get a knowlegable opinion of his scheme...
> whch is probably already well known to astronomical experts
> somwhere.
>[dkelvey]
>Hi
> The problem with this method was that the Moon was not always
>seen at night. Ships can't wait for the Moon.
> Using a clock, one could use Moon, Sun or stars.
>Dwight
[Hammond]
   Crossing the Atlantic or Pacific took months in those
days.  They certainly had plenty of time to wait for
favorable observing conditions.
   Besides, in those days there was no way of even
determining longitude on land!  The longitude of the
American colonies could only be approximately guessed for
instance.  Surely Swedenborg's method could have been used
to determine the longitude of Boston, New York or St.
Augustine.
     SCIENTIFIC PROOF OF GOD WEBSITE
 http://geocities.com/scientific_proof_of_god
   mirror site:
 http://proof-of-god.freewebsitehosting.com
      GOD=G_uv   (a folk song on mp3)
 http://interrobang.jwgh.org/songs/hammond.mp3
===
Subject: Re: SWEDENBORG ON LONGITUDE
        posting-account=jPnQ2goAAAA461y3QD0lbyw0oKeThma1
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Safari/525.20,gzip(gfe),gzip(gfe)
the moon can always be seen at some time of night, at sea,
except for a day or two per orbit.  that is the basis
of Captain Maui's method in Erastosthenes' expedition,
which got as far as Chile (and perhaps no further;
see www.wlym.com).
re Swedenborg,
can you lift The Urantia Book?...
can God launch an albatross that he can't float?
> æThe problem with this method was that the Moon was not 
always
> seen at night. Ships can't wait for the Moon.
> æUsing a clock, one could use Moon, Sun or stars.
thus:
of course, even if you can see Venus, some times,
carefully shielding your eyes from the bright horizon
of dawn or dusk, why would that mean that the camera will see it?
thus:
is that purposeful?
speaking of whether you're really a wigwam or a teepee [*],
Islam has many important cultural aspects, perhaps all
of them simply being codifications of Arabic ones, but
I don't plan on converting. I mean, I'd love
to be a camel jockey, two, but Obama's just-announced conversion
to the anticonstitutional faith-based initiatives [**], really,
has me wondering about camel poop ... and other forms
of alternative energy.  I only just read that
Obama shills for corn ethanol.
there is no separation of church & state,
for crying out loud; doesn't the DNC know this?
* the trouble with you is,
you're two tents; ha.
** here, we can perhaps see the true meaning
of *antidisestablishmentarianism*, if it's the case that
the Bill of Rights is disestablishmentarian with regard
to religions.  I'd always thought, it was just a nonsequiter.
>search:
>buckminster june-26-2008 OR 6.26.2008
which one of you guys has ever photographed foreground objects,
against a supposedly starry background?...  well,
I haven't, either, but I wouldn't even try.  I mean,
do you ever see objects in the foreground, regardless of focus,
in astronomical photography?
this is a God-am Photo One dot One lesson -- yeesh!
> well, how about, what maximum angle over the horizon,
> have you ever seen Venus without any visual aids?
> * the trouble with you is,
> you're two tents; ha.
--Seargent Cheeny Pepper,
Give war a chance in the Sudan, Rhodesia, and
other former colonial markets!
===
Subject: Re: SWEDENBORG ON LONGITUDE
the moon can always be seen at some time of night, at sea,
except for a day or two per orbit.
Babbling cretin. Not only are you idiotically wrong about
night, at sea has no relevance whatsoever.
===
Subject: Re: SWEDENBORG ON LONGITUDE
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?
Something similar happened with Isaac Newton in his later years.
===
Subject: Re: SWEDENBORG ON LONGITUDE
>[Hammond]
>              SWEDENBORG ON LONGITUDE
>    In 1721 (Baron) Emanuel Swedenborg published a simple
> method for finding longitude from the moon.  It is known
> that he personally discussed it in London with Royal
> Astronomer Flamsteed and Edmond Halley.  Apparently neither
> of them thought it erroneous or incompetent.
>    Swedenborg of course later became world famous for
> publishing visionary theological speculations.  According to
> me he had a schizophrenic reaction at the age of 55 and gave
> up routine science to conduct the world's first scientific
> investigation of schizophrenic visions, and which resulted
> in his celebrated book _Heaven and Hell_; the world's most
> famous quasi-technical description of Life After Death.
>    What I want to know, is how brilliant a scientist and
> mathematician was he before he became fascinated with mental
> phenomena?
[Virgi]
>Something similar happened with Isaac Newton in his later years.
>
[Hammond]
   Wolfgang Pauli too.  In fact Carl Jung treated Pauli for
mild mental disturbance after he married a stripper and had
a drunken driving accident.  In 1952 they jointly published
a book entitled _The Interpretation of Nature and the
Psyche_ in which they proposed their own watered down
Swedenborgian verison of the physics of mental phenomena.
   Fact is many scientists get slightly schizo in their old
age.  Too much studying and too little sex.  Swedenborg was
simply the first to take a high powered scientific look at
the contents of mental disturbance.  Pauli was second.  Now
Hammond is the third.  This progression finally led Hammond
to prove that  GOD=G_uv  which proves that God is an actual
curvature of (subjective) spacetime.
   Hammond has discovered the scientific explanation of
God, but I am still working on the problem of Life After
Death.  Which is why I am currently investigating Swedenborg
in depth.
     SCIENTIFIC PROOF OF GOD WEBSITE
 http://geocities.com/scientific_proof_of_god
   mirror site:
 http://proof-of-god.freewebsitehosting.com
      GOD=G_uv   (a folk song on mp3)
 http://interrobang.jwgh.org/songs/hammond.mp3
===
Subject: solutions manual to many textbook
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I am a solutions manual collector, I offer solutions manual ebook
servicesNote:
all solutions manual in soft copy
that mean in Adobe Acrobat Reader  (PDF ) format. if you want any book
not just solutions just contact with us to get the  solution manual
you want please send message to sendsolutions@hotmail.com
sendsolutions(at)hotmail.com replace (at) to @ ,please email to
me .This is my part of solutions manual list ,If you want any other
solutions manual which is not in my solutions list,don.81ft give
up .please email to sendsolutions(at)hotmail.com
Calculus A Complete Course 6th Edition by R.A. Adams
Cisco Technical Solution Series: IP Telephony Solution Guide Version
2.0
Control Systems 4th edition by Norman S. Nise
Corporate Finance, Second Edition by Peter Bossaerts
Electrical Circuits  6th           by James W. Nilsson, and Susan A.
Electrical Circuits  7th            by James W. Nilsson, and Susan A.
Elements of Electromagnetics  3rd   by  Sadiku
Elements of engineering electromagnetics  (6/e)      by  N.N.RAO
Engineering electromagnetics (6/e)                   by  HAYT
Engineering electromagnetics (7/e)                  by  HAYT
Fundamentals of Signals and systems using web and matlab third edition
by deepa08
Introduction to Mathematical Statistics 6/E   Robert V. Hogg
Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang
Modern Digital and Analog Communication Systems - B.P. Lathi 3rd
Edition Solution Manual
Modern Digital and Analog Communication Systems (3/e)
Numerical methods for engineers 5th  by Chapra
Physics of the Solar Corona: An Introduction with Problems and
Solutions by Markus Aschwanden
Physics, 5th Edition by Halliday, Resnick, and Krane .81iThe Internet
Short Edition.81j
Based on the BWF-HHMICourse in Scientif?c Management for theBeginning
Academic Investigator
Principles of Highway Engineering and Traffic Analysis, 3e,by Fred L.
Mannering, Walter P. Kilareski, Mathematics for Economists-Carl P.
Simon Lawrence Blume Solution Manual
Probability and Statistics for Engineering and the Sciences 7/e JAY
L.DEVORE
Separation Process Principles, 2nd Ed.,                   by  Seader,
Henley
Signal Processing First-Mclellan, Schafer & Yoder Solution Manual
Signals and Systems (2nd Edition)    By Oppenheim, Willsky and Nawab
STATISTICAL METHODS FOR CRIMINOLOGY AND CRIMINAL JUSTICE
Thermodynamics: An Engineering Approach,6th          by Cengel
Two-Dimensional Incompressible Navier-Stokes Equations- Maciej Matyka
University Physics with Modern Physics:,11 th  By Hugh D. Young, Roger
A. Freedman,
University Physics with Modern Physics:,12th  By Hugh D. Young, Roger
A. Freedman,
===
Subject: Re: solutions manual to many textbook
        posting-account=qL0m4woAAAAXbWR0iVY3aIl0MbFVyU9L
        Gecko/2008061015 Firefox/3.0,gzip(gfe),gzip(gfe)
> I am a solutions manual collector, I offer solutions manual ebook
> servicesNote:
> all solutions manual in soft copy
> that mean in Adobe Acrobat Reader  (PDF ) format. if you want any book
> not just solutions just contact with us to get the  solution manual
> you want please send message to sendsoluti...@hotmail.com
> sendsolutions(at)hotmail.com replace (at) to @ ,please email to
> me .This is my part of solutions manual list ,If you want any other
> solutions manual which is not in my solutions list,don't give
> up .please email to sendsolutions(at)hotmail.com
> Calculus A Complete Course 6th Edition by R.A. Adams
> Cisco Technical Solution Series: IP Telephony Solution Guide Version
> 2.0
> Control Systems 4th edition by Norman S. Nise
> Corporate Finance, Second Edition by Peter Bossaerts
> Electrical Circuits  6th           by James W. Nilsson, and Susan A.
> Electrical Circuits  7th            by James W. Nilsson, and Susan A.
> Elements of Electromagnetics  3rd   by  Sadiku
> Elements of engineering electromagnetics  (6/e)      by  N.N.RAO
> Engineering electromagnetics (6/e)                   by  HAYT
> Engineering electromagnetics (7/e)                  by  HAYT
> Fundamentals of Signals and systems using web and matlab third edition
> by deepa08
> Introduction to Mathematical Statistics 6/E   Robert V. Hogg
> Microelectronics I & II 1st edition by by Dr. Wen-Ching Chang
> Modern Digital and Analog Communication Systems - B.P. Lathi 3rd
> Edition Solution Manual
> Modern Digital and Analog Communication Systems (3/e)
> Numerical methods for engineers 5th  by Chapra
> Physics of the Solar Corona: An Introduction with Problems and
> Solutions by Markus Aschwanden
> Physics, 5th Edition by Halliday, Resnick, and Krane .81iThe Internet
> Short Edition.81j
> Based on the BWF-HHMICourse in Scientif?c Management for theBeginning
> Academic Investigator
> Principles of Highway Engineering and Traffic Analysis, 3e,by Fred L.
> Mannering, Walter P. Kilareski, Mathematics for Economists-Carl P.
> Simon Lawrence Blume Solution Manual
> Probability and Statistics for Engineering and the Sciences 7/e JAY
> L.DEVORE
> Separation Process Principles, 2nd Ed.,                   by  Seader,
> Henley
> Signal Processing First-Mclellan, Schafer & Yoder Solution Manual
> Signals and Systems (2nd Edition)    By Oppenheim, Willsky and Nawab
> STATISTICAL METHODS FOR CRIMINOLOGY AND CRIMINAL JUSTICE
> Thermodynamics: An Engineering Approach,6th          by Cengel
> Two-Dimensional Incompressible Navier-Stokes Equations- Maciej Matyka
> University Physics with Modern Physics:,11 th  By Hugh D. Young, Roger
> A. Freedman,
> University Physics with Modern Physics:,12th  By Hugh D. Young, Roger
> A. Freedman,
What's this? An selling advertisement?
===
Subject: Simulation of a non-deterministic Turing machine
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Hello.
why is it possible to simulate a non-deterministic turing machine on a
deterministic one in exponential time? I can't see that. Can anybody
help me?
S.
===
Subject: Re: Simulation of a non-deterministic Turing machine
> why is it possible to simulate a non-deterministic turing machine on a
> deterministic one in exponential time? I can't see that. Can anybody
> help me?
The usual model of a non-deterministic turing machine has
non-determinism in the form of choosing a 0 or a 1 in some of its
steps. If the machine runs at most n steps in all possible runs, then
there are at most 2^n different runs, with at most n steps each, that
is all runs together have at most n*2^n steps. If there are any
infinite runs, then simulation is not sensible, anyway. Now you can be
sure that there are only exponentially many different states to look
at.
The key point is, that the non-determinism is bounded, the choice is
only from a finite set (here 0 or 1). If you take unbounded
non-determiniscm, e.g. take any natural number in every step, then the
exponential-time simulation will fail, because the number of different
runs becomes uncountable (another diagonalization argument).
The remaining problems are purely technical, howto restart a turing
machine with the same input, and howto handle the nondeterminism, i.e.
make sure you have looked at all possible sequences of choices. I think
there are other postings explaining that nicely.
Bernd Strieder
===
Subject: Re: Simulation of a non-deterministic Turing machine
> Hello.
 why is it possible to simulate a non-deterministic turing machine on a
> deterministic one in exponential time? I can't see that. Can anybody
> help me?
 S.
A standard way is to use a 3-tape DTM:
Tape 1 consists of the input to the NTM. It is write-only.
Tape 2 contains the working tape for the NTM being simulated. It
initially contains a copy of Tape 1.
Tape 3 is the heart of the simulation. It contains a list of the
current states of the NTM, in the following sense: Assume that
the NTM has at most k possible nondeterministic moves from any
state and encode a move using a string over the alphabet {1, ... , k},
so, for example, the string 1322 would represent the moves the
NTM would make by choosing alternative 1 as the first move, followed
by option 3 from that state's possible second moves, and so on.
Generate the move strings lexicographically and for each choice
simulate (using Tape 2) the NTM's action on that string of moves.
There are some details missing here, of course, but the idea should
be clear enough. Essentially, you are performing a breadth-first
search through the k-way tree of possible moves, which will obviously
take time O(k^n) to simulate n moves of the NTM.
Rick
(BTW, a good forum for questions like this is the comp.theory NG.)
===
Subject: Topology with quotient map R_K.
Hello teacher~
Let K denote the set of all numbers of the form 1/n, for n in Z+,
and B be the collection of all open intervals (a, b), along with
all sets of the form (a, b) - K.
The topology generated by B will be called the K-topology on R.
R_K denotes the real line in the K-topology.
Let Y be the quotient space obtained from R_K by collapsing
the set K to a point;
let p : R_K -> Y be the quotient map.
(1) Show that Y satisfies the T_1 axiom, but is not Hausdorff.
-------------------------------------------------------------
Suppose that Y is T_2.
By the property of partition of R_K,(one class maps a element of Y.)
Let p(K) = y_K  in Y.
Let C is a class containing 0 of the partition with R_K.
Let p(C) = y_0   in Y.
Since Y is T_2,
For y_K ,  y_0  in Y,  (y_1 =/= y_2)
there is a open set V_K containing y_K  and
there is a open set V_0 containing y_0.
(V_K , V_0 are two disjoint element of quotient topology with Y).
so, [p^{-1}(V_K)]  /  [p^{-1}(V_0)]  = empty set
and K  subset  p^{-1}(V_K)
and {0} subset  p^{-1}(V_0).
and p^{-1}(V_K) and p^{-1}(V_0) are disjoint open set.
But {0} in K'.
so, [p^{-1}(V_0)] / K =/= emtpy.
so, p^{-1}(V_K) and p^{-1}(V_0) are not disjoint.
so, contradiction.
---------------------------------------------------
I don't know about T_1 well.
so, I need your advice. 
===
Subject: Re: Topology with quotient map R_K.
> Let K denote the set of all numbers of the form 1/n, for n in Z+,
> and B be the collection of all open intervals (a, b), along with
> all sets of the form (a, b) - K.
> The topology generated by B will be called the K-topology on R.
>
This is the same as the topology for R with RK as an added open set.
> R_K denotes the real line in the K-topology.
> Let Y be the quotient space obtained from R_K by collapsing
> the set K to a point;
> let p : R_K -> Y be the quotient map.
 (1) Show that Y satisfies the T_1 axiom, but is not Hausdorff.
>
Let p(K) = q in Y.  If Y is T1, then {q} is closed.
Hence p^-1(q) = K is closed, which it isn't.
Exercise.  R_K is T0.
----
===
Subject: Re: Topology with quotient map R_K.
> Let K denote the set of all numbers of the form 1/n, for n in Z+,
> and B be the collection of all open intervals (a, b), along with
> all sets of the form (a, b) - K.
> The topology generated by B will be called the K-topology on R.
> This is the same as the topology for R with RK as an added open set.
> R_K denotes the real line in the K-topology.
> Let Y be the quotient space obtained from R_K by collapsing
> the set K to a point;
> let p : R_K -> Y be the quotient map.
> (1) Show that Y satisfies the T_1 axiom, but is not Hausdorff.
> Let p(K) = q in Y.  If Y is T1, then {q} is closed.
> Hence p^-1(q) = K is closed, which it isn't.
Maybe... wrong problem.
Anyway, this is a problem of Munkres book. 
===
Subject: Help with Calculus of Variations Problem